Logic & Word Problemsageswork-backwardsum-constraint
Anna and Bella are celebrating their birthdays together. Five years ago, when Bella turned 6 years old, she received a newborn kitten as a birthday present. Today the sum of the ages of the two children and the kitten is 30. How many years older than Bella is Anna?
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Answer: C — 3 years older.
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Hint 1 of 2
The two ages you can actually pin down are Bella's and the kitten's — nail those today, and the total does the rest.
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Hint 2 of 2
Bella was 6 five years ago, so she's 11; the kitten was newborn, so it's 5. Subtract both from 30 to uncover Anna's age, then compare.
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Approach: pin down the ages you can know first, then let the total reveal the unknown
Insight: the kitten is the easy clue, not a distraction — “newborn five years ago” means it's exactly 5 today. And Bella, 6 five years ago, is 11 today. Both ages are now fixed.
Anna is the only mystery, and the total 30 hands her to us: Anna = 30 − 11 (Bella) − 5 (kitten) = 14.
Students Arn, Bob, Cyd, Dan, Eve, and Fon are arranged in that order in a circle. They start counting: Arn first, then Bob, and so forth. When the number contains a 7 as a digit (such as 47) or is a multiple of 7 that person leaves the circle and the counting continues. Who is the last one present in the circle?
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Answer: D — Dan.
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Hint 1 of 2
The work isn't in the counting — it's in writing the "danger" numbers first: 7, 14, 17, 21, 27… Once you have that short list, you only care about who says those numbers, not the boring ones in between.
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Hint 2 of 2
The technique is a simulation: keep a shrinking list of who's still in, and remember the count keeps climbing while the people loop — restart from whoever is next, never from 1.
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Approach: step through the counting, eliminating people
Step 1 is the setup that makes everything easy: list the eliminating counts in order — 7, 14, 17, 21, 27, … (multiples of 7, or any number containing the digit 7). Now you only have to find who says each of these.
You'll see it again: "person leaves the circle, counting continues" problems are simulations — resist hunting for a formula and just bookkeep carefully; the only trap is forgetting the count keeps rising as people drop out.
The ten-letter code "BEST OF LUCK" represents the ten digits 0–9, in order. What 4-digit number is represented by the code word "CLUE"?
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Answer: A — 8671.
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Hint 1 of 2
The code is a one-to-one key: the 1st letter stands for 0, not 1 — counting starts at zero.
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Hint 2 of 2
Build the lookup table once, then read off the answer; this is just decoding with a substitution cipher.
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Approach: map letter position to digit
Number the letters from 0: B=0, E=1, S=2, T=3, O=4, F=5, L=6, U=7, C=8, K=9. The trap is starting at 1 — the ten letters cover 0–9, so the first letter is 0.
Read C-L-U-E off the table: 8-6-7-1 = 8671.
Shortcut: the choices already start 8… or 9…, so finding just C=8 eliminates two answers instantly.
To complete the grid below, each of the digits 1 through 4 must occur once in each row and once in each column. What number will occupy the lower right-hand square?
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Answer: B — 2.
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Hint 1 of 2
The question asks about one cell, so don't solve the whole puzzle — only follow the row and column that pass through that corner.
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Hint 2 of 2
Sudoku logic: attack the most-filled line first. A cell forbids every digit already in its row or column, so the tightest spots get forced one at a time, and each forced cell tightens the next.
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Approach: force the target cell using only its own column
Aim at column 4, the column through the corner we want. It already holds a 4 (row 3), so no other cell in it can be 4 — that's the lever.
Row 1's two blanks are 3 and 4, but column 4 has banned 4, so row 1 col 4 = 3. Same move on row 2 (blanks 1 and 4): col 4 = 1.
Column 4 now shows 3, 1, 4, leaving only one digit for the bottom-right square: 2.
Transfers everywhere: in any Latin-square / Sudoku cell, when three of four values are accounted for in a line, the fourth is forced — chase eliminations, not the whole grid.
A five-legged Martian has a drawer full of socks, each of which is red, white or blue, and there are at least five socks of each color. The Martian pulls out one sock at a time without looking. How many socks must the Martian remove from the drawer to be certain there will be 5 socks of the same color?
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Answer: D — 13.
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Hint 1 of 2
'To be certain' means plan for the unluckiest draw possible. Ask: what's the most socks you could hold and still not have 5 of any one color?
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Hint 2 of 2
This is the pigeonhole idea: pile each color as high as it can go without hitting the target (4 each), then the very next sock is forced to push some color over.
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Approach: pigeonhole — build the worst case, then add one
Imagine the meanest possible luck: 4 red, 4 white, 4 blue. That's 12 socks and still no color has reached 5.
There's nowhere left to hide — the 13th sock must be a 4th color's... no, must be the 5th of some color. So 13 guarantees it.
Why this transfers: for 'how many to guarantee' problems, find the largest haul that fails the goal (here 3×4 = 12), then add 1. The five legs in the story are pure distraction — only the three colors matter.
Ms. Hamilton's eighth-grade class wants to participate in the annual three-person-team basketball tournament. The losing team of each game is eliminated from the tournament. If sixteen teams compete, how many games will be played to determine the winner?
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Answer: D — 15 games.
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Hint 1 of 2
Don't try to draw the bracket round by round. Flip the question: instead of 'how many games?', ask 'how many teams must disappear?' Each game removes exactly one team.
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Hint 2 of 2
This is the single-elimination principle: to crown one champion out of N teams, exactly N − 1 teams must lose — and each game produces exactly one loser, so games = N − 1. The bracket's shape never matters.
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Approach: count losers, not games
Every game knocks out exactly one team. The tournament ends with 1 champion, so 16 − 1 = 15 teams had to be eliminated.
One game per elimination ⇒ 15 games — no bracket-drawing needed.
Worth keeping: any 'games to find a single winner in single-elimination' = (teams − 1). A 64-team bracket takes 63 games; a 128-team draw takes 127. The answer never depends on byes or how the rounds line up.
Another way — add up the rounds (the long way):
16 teams → 8 games → 8 teams → 4 games → 4 teams → 2 games → 2 teams → 1 game.
8 + 4 + 2 + 1 = 15. Same answer — and notice it equals 16 − 1, confirming the shortcut.
Logic & Word Problemswork-backwardcareful-counting
Each principal of Lincoln High School serves exactly one 3-year term. What is the maximum number of principals this school could have during an 8-year period?
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Answer: C — 4.
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Hint 1 of 2
8 ÷ 3 isn't a whole number, so the terms don't line up neatly with the window. The trick to MAXIMIZING is to let the window catch the tail end of one term and the front end of another — waste partial terms at both edges.
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Hint 2 of 2
To maximize how many things overlap a fixed window, push a boundary just inside each end. The two end-principals only need to touch the window by a single year.
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Approach: let partial terms hang over both ends of the window
Imagine year 1 is the *final* year of some principal's term — that principal counts, even though most of their term was before our window.
Their 3-year terms then cover years 2–4 (principal 2) and 5–7 (principal 3). That uses years 2 through 7.
Year 8 is the *first* year of a fourth principal — they count too. Total: 4 principals.
The principle: to fit the most fixed-length blocks into a window, align a block boundary just inside each end so the two end blocks only barely overlap — you gain a partial block at each edge. (Choice E's '8' is the trap: a 3-year term can never be just 1 year.)
Bo, Coe, Flo, Jo, and Moe have different amounts of money. Neither Jo nor Bo has as much money as Flo. Both Bo and Coe have more than Moe. Jo has more than Moe, but less than Bo. Who has the least amount of money?
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Answer: E — Moe.
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Hint 1 of 2
The question only asks for the least, so you don't need to sort all five — you just need the one person nobody falls below. Watch which name keeps showing up on the small side of every comparison.
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Hint 2 of 2
Scan the clues for the name that is always "more than" someone else. The person who never appears as the bigger one is your answer.
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Approach: look only for the bottom, not the full order
Sweep the clues for who loses each comparison. Moe is beaten by Bo, by Coe, and by Jo directly. And since Flo > Bo > Moe, Flo beats Moe too.
Every other person sits above Moe, so Moe has the least.
Why this transfers: for a "who is least/greatest" question you only need the extreme, not the whole ranking — find the name that's on the losing side of every clue and stop. Chaining Flo > Bo > Moe shows you can hop across people you were never directly told to compare.
Don't guess the top circle — attack the most constrained edge first. With only the digits 1–6, which edge-sum can be made in only one way?
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Hint 2 of 2
The edge labeled 10 is the tightest: among 1–6, the only pair adding to 10 is 4 + 6. Pin those two circles down, then each neighboring sum forces the next digit like dominoes.
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Approach: start at the most-constrained edge, then let the sums force each digit
Scan for the edge with the fewest options. The sum 10 is the giveaway: out of 1–6, only 4 + 6 reaches 10, so that left edge's two circles are 4 and 6 in some order.
Test which way round: the edge above it sums to 9. If the upper circle is 4, the top is 9 − 4 = 5; the lower one is then 6. (The other way, top = 9 − 6 = 3, collides later.)
Now the dominoes fall: bottom-left 6 with edge 8 → bottom-middle 2; edge 5 → bottom-right 3; right edge 4 → upper-right 1; and the last check 5 + 1 = 6 matches the top-right edge. Digits used: {5,4,6,2,3,1} = 1–6 exactly once. ✓
So the top circle is 5.
Why this transfers: in any fill-the-grid / fill-the-graph puzzle, begin at the cell with the fewest legal choices — one forced value usually triggers a chain that solves the rest with no guessing.
Don't try to trace whole routes — too many. Instead label each town with its OWN shortest distance from A, in the order the towns can be reached.
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Hint 2 of 2
Technique (shortest-path relaxation): a town's best distance = the smallest "(best distance to a town that feeds it) + (that edge)". Solve them in flow order A, X, M, Y, C, Z.
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Approach: shortest-path table, town by town
Rather than list every full route, find the shortest distance to each town from A, building up in the order towns become reachable. Each town's value = the cheapest "(arrived-distance to a feeder) + (its edge to here)." Start: A→X = 5 (only way in).
A→M = min(8 direct, 5 + 2 via X) = 7. Going through X beats the direct road.
A→Y = min(5 + 10 via X, 7 + 6 via M) = 13.
A→C = min(7 + 14 via M, 13 + 5 via Y) = 18.
A→Z = min(7 + 25 via M, 13 + 17 via Y, 18 + 10 via C) = 28 (via C). This transfers: this is Dijkstra's idea — once a town has its final shortest label, every later town can lean on it, so you never re-explore whole paths.
In a tournament there are six teams that play each other twice. A team earns 3 points for a win, 1 point for a draw, and 0 points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?
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Answer: C — 24 points each.
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Hint 1 of 2
To push the top three as high as possible, give them every point that's "free" — let all three crush the bottom three in every game. The only points in question are the ones they fight over among themselves.
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Hint 2 of 2
A win-then-loss split between two teams yields 3 + 0 = 3 points total, but a draw-draw yields only 1 + 1 = 2. So to maximize and keep the trio tied, settle their head-to-heads with decisive wins, not draws.
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Approach: sweep the bottom, split decisively among the tied top
Each top team plays the bottom three twice each = 6 games. Winning all of them gives 6 × 3 = 18 points — the maximum any team can grab from outside the trio.
Within the top three, the three pairs play twice each. To keep all three level, give each pair a 1-win, 1-loss split. A win is worth 3 and a loss 0 (better than two draws at 1 each), and the splits cancel out so the trio stays tied.
Each top team sits in 2 of those pairs and wins one game in each: +3 + 3 = 6 more.
Maximum each = 18 + 6 = 24.
Why this transfers: in "maximize a tied group" problems, hand the group all the outside wins, then realize that 3-for-a-win beats splitting points via draws — symmetry keeps the tie while decisive results keep the total high.
Peter, Emma, and Kyler played chess with each other. Peter won 4 games and lost 2 games. Emma won 3 games and lost 3 games. If Kyler lost 3 games, how many games did he win?
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Answer: B — 1 win.
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Hint 1 of 2
You're not told how many games were played — and you don't need to be. Every single game produces exactly one winner and one loser, so across everyone, total wins must equal total losses.
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Hint 2 of 2
Counting principle: when each event creates one of each kind, the two totals are forced equal. Add up the losses, set wins to match, solve for the unknown.
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Approach: total wins = total losses
Each game makes one winner and one loser, so summed over all three players, wins = losses. This balance lets you skip working out who played whom.
Total losses: 2 + 3 + 3 = 8. So total wins must also be 8: 4 + 3 + Kyler = 8.
Kyler's wins = 8 − 7 = 1.
Why this transfers: any tournament/handshake/edge-counting setup where each event contributes equally to two tallies lets you equate the tallies instead of reconstructing the schedule.
In an All-Area track meet, 216 sprinters enter a 100-meter dash competition. The track has 6 lanes, so only 6 sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?
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Answer: C — 43 races.
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Hint 1 of 2
Flip the question. Instead of asking "how many races happen?" (which means tracking winners through messy rounds), ask "how many sprinters must DISAPPEAR?" That number is fixed, no matter how the bracket is arranged.
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Hint 2 of 2
This is the single-elimination principle: to crown ONE champion out of N, exactly N − 1 competitors must be eliminated — and here each race removes a fixed 5 of them. So races = (eliminations needed) ÷ (eliminated per race).
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Approach: count what must be eliminated, not the races themselves
There's exactly one champion, so 216 − 1 = 215 sprinters must be eliminated — this total never changes, however the heats are organized.
Each race eliminates exactly 5 (the non-winners), so the number of races = 215 ÷ 5 = 43.
Worth keeping: any "games needed to find one winner" equals (players − 1) ÷ (losers per game). A 64-team single-elimination bracket needs 64 − 1 = 63 games, no bracket-drawing required.
Sanity check: 215 isn't a multiple of 5? It is (215 = 5 × 43), and the final race must end with 1 winner from however many remain — the arithmetic working out cleanly confirms the setup.
Three members of the Euclid Middle School girls' softball team had the following conversation. Ashley: I just realized that our uniform numbers are all 2-digit primes. Bethany: And the sum of your two uniform numbers is the date of my birthday earlier this month. Caitlin: That's funny. The sum of your two uniform numbers is the date of my birthday later this month. Ashley: And the sum of your two uniform numbers is today's date. What number does Caitlin wear?
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Answer: A — 11.
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Hint 1 of 3
The dates do double duty as ordering clues: every pairwise sum is a calendar date (1–31), and the words "earlier" / "later" / "today" rank those three sums smallest < middle < largest.
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Hint 2 of 3
Sums ≤ 31 squeeze the three primes down to a tiny list — find a triple of two-digit primes whose three pairwise sums all fit and are distinct.
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Hint 3 of 3
Each girl names the sum of the OTHER two. So the girl with the largest sum-of-others must herself be the smallest number — the big two are added without her.
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Approach: shrink the prime list, then use date order to rank the sums
Each pairwise sum is a date ≤ 31. Two-digit primes are 11, 13, 17, 19, 23, 29, but any pair using 23 or 29 overshoots 31, so the three numbers come from {11, 13, 17, 19}.
Pick a triple with three distinct sums: {11, 13, 17} gives 11+13 = 24, 11+17 = 28, 13+17 = 30 — all distinct and ≤ 31. ✓
Order the clues: Bethany's date (earlier) is the smallest sum 24, today is the middle 28, Caitlin's date (later) is the largest 30. Each girl quotes the sum of the OTHER two.
Caitlin's date = 30 = sum of the other two = Ashley + Bethany, so Ashley and Bethany are the two larger numbers {13, 17}, leaving Caitlin = 11.
Cross-check: Bethany's date 24 = Ashley + Caitlin = Ashley + 11 ⇒ Ashley = 13, so Bethany = 17, and today = Ashley + Bethany... wait, today = Bethany + Caitlin = 17 + 11 = 28. ✓ All three statements hold.
Caitlin wears 11.
Why this transfers: when each clue references "the others," the person tied to the biggest total is the one left out of it — so largest sum-of-others ↔ smallest own value. Spotting that inverse ordering cracks many "sum of the other two" puzzles.
One day the Beverage Barn sold 252 cans of soda to 100 customers, and every customer bought at least one can of soda. What is the maximum possible median number of cans of soda bought per customer on that day?
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Answer: C — 3.5.
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Hint 1 of 3
The median of 100 sorted values only looks at the 50th and 51st. Everything below the 50th is dead weight you control — so spend as few cans as possible there to free up cans for the middle.
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Hint 2 of 3
"At least one each" means the floor is 1, so set the bottom 49 customers to 1 can. That's the cheapest legal way to clear them out of the way.
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Hint 3 of 3
Now push the 50th and 51st as high as the leftover cans allow, remembering every customer above the 50th must be ≥ the 50th's value.
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Approach: minimize the first 49, push the median pair as high as possible
The median of 100 sorted counts is the average of the 50th and 51st. To maximize it, minimize everyone below: set customers 1–49 to the legal minimum of 1 can each (uses 49 cans, leaves 252 − 49 = 203 for the top 51).
Let the 50th count be a and the 51st be b with a ≤ b. To spend the fewest cans above the median, make customers 51–100 all equal to b. Then the last 51 use a + 50b ≤ 203.
Push b up. b = 4: need a + 200 ≤ 203 ⇒ a ≤ 3, so take a = 3 (and 3 ≤ 4 ✓). Median = (3 + 4)/2 = 3.5.
b = 5 would need a + 250 ≤ 203 — impossible, so 3.5 is the ceiling.
Why this transfers: to maximize a median, starve the bottom half down to the minimum so the saved resource can lift the two middle values — the same "rob the cheap end to fund the position you care about" trick appears all over optimization problems.
A certain calculator has only two keys [+1] and [×2]. When you press one of the keys, the calculator automatically displays the result. For instance, if the calculator originally displayed "9" and you pressed [+1], it would display "10." If you then pressed [×2], it would display "20." Starting with the display "1," what is the fewest number of keystrokes you would need to reach "200"?
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Answer: B — 9 keystrokes.
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Hint 1 of 2
Going forward from 1 you'd face two choices at every step — a branching mess. Going backward from 200, each step is forced, so there's nothing to guess.
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Hint 2 of 2
Reverse the keys: undoing ×2 is ÷2, undoing +1 is −1. At an even number, halving is the bigger leap toward 1, so prefer it; only subtract 1 when the number is odd and you have no choice.
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Approach: run the keys in reverse from 200 to 1
Work down from the target. Halve whenever you can (the powerful move), and subtract 1 only when forced by an odd number:
That's 9 steps, and each reverse step is one forward keystroke.
Why backward is optimal here: at an even number, halving always reaches 1 in fewer moves than peeling off 1's, and at an odd number you're forced to subtract 1 anyway — so this greedy reverse path can't be beaten. Working from the goal is the go-to trick whenever the forward direction branches but the backward one is determined.
Five friends compete in a dart-throwing contest. Each one has two darts to throw at the same circular target, and each individual's score is the sum of the scores in the target regions that are hit. The scores for the target regions are the whole numbers 1 through 10. Each throw hits the target in a region with a different value. The scores are: Alice 16, Ben 4, Cindy 7, Dave 11, Ellen 17. Who hits the region worth 6 points?
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Answer: A — Alice.
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Hint 1 of 2
Every region 1–10 is hit exactly once across all ten darts — so once a number is 'claimed', it's gone for everyone else. Attack the score with the fewest possible pairs first: that's the smallest total, Ben = 4.
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Hint 2 of 2
The technique is most-constrained-first (like solving a Sudoku): handle the score that has only one possible pair, lock those numbers away, and the next score usually collapses to a single option too — a chain reaction.
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Approach: most-constrained-first chain
Ben = 4 with two different regions: the only option is 1 + 3 (2 + 2 isn't allowed). Lock away 1 and 3.
Cindy = 7 could be 1+6, 2+5, or 3+4 — but 1 and 3 are now gone, leaving only 2 + 5. Lock away 2 and 5.
Dave = 11 from what's left {4,6,7,8,9,10}: only 4 + 7 survives (5+6 and 2+9 use claimed numbers). Lock away 4 and 7.
Alice and Ellen must split the remaining {6, 8, 9, 10}. Alice = 16 forces 6 + 10 (since 7+9 uses a gone 7), and Ellen = 17 = 8 + 9.
So Alice is the one who hits 6.
Why start small: the smallest and largest totals have the fewest valid pairs, so they're the safest place to begin — each forced pair removes numbers and tightens the next one. You'll use this 'pin the most-restricted thing first' habit in every logic-grid puzzle.
In this addition problem, each letter stands for a different digit.
T W O
+ T W O
-------
F O U R
If T = 7 and the letter O represents an even number, what is the only possible value for W?
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Answer: D — W = 3.
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Hint 1 of 2
TWO + TWO is just TWO doubled, so every column is a digit being doubled (maybe plus a carry) — that's a strong constraint.
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Hint 2 of 2
Start at the most-constrained column. With T = 7, the hundreds give 14 (+ carry), forcing the leading F = 1 and pinning O; then "O is even" finishes it.
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Approach: double TWO column by column, starting where it's pinned
TWO + TWO = 2 × TWO, and doubling a 3-digit number can only push it to 4 digits by carrying a 1 — so F = 1, the only digit a carry can produce.
Hundreds column is the most pinned: 7 + 7 = 14, plus a possible carry from the tens, lands the hundreds digit O at 4 or 5. Since O is even, O = 4 — and that means nothing carried in, so the tens column did not overflow.
Units: O + O = 4 + 4 = 8, so R = 8 with no carry. The tens column is therefore just W + W = U, with no carry either way.
Now place W: 2W = U must stay under 10 (no overflow) and U must avoid the used digits 7, 1, 4, 8. W = 0→U = 0 (clashes with W and is reused), W = 1, 4 taken, W = 2→U = 4 (taken). Only W = 3 works, giving U = 6. W = 3.
You'll see this again: in cryptarithms, attack the column with the fewest unknowns first (often the leading carry, which is almost always 1), and let each forced digit cascade into the next column.
The six children listed below are from two families of three siblings each. Each child has blue or brown eyes and black or blond hair. Children from the same family have at least one of these characteristics in common. Which two children are Jim's siblings?
Child
Eye Color
Hair Color
Benjamin
Blue
Black
Jim
Brown
Blond
Nadeen
Brown
Black
Austin
Blue
Blond
Tevyn
Blue
Black
Sue
Blue
Blond
Show answer
Answer: E — Austin and Sue.
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Hint 1 of 2
A sibling of Jim must MATCH Jim on eyes or hair — cross off anyone who shares neither of his traits.
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Hint 2 of 2
Look for three children who share a single trait among themselves; if they make one valid family, the leftovers must be the other.
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Approach: shortlist by Jim's traits, then split the six into two valid families
Jim has brown eyes and blond hair. A sibling must match at least one, so anyone with blue eyes and black hair is out. That leaves only Nadeen (brown eyes), Austin (blond), and Sue (blond) as possible siblings — and Jim has exactly 2 siblings, so two of these three are his.
Now use the "other family" to break the tie. Benjamin, Nadeen, and Tevyn all have black hair — they form a complete, valid family of three on their own. That uses up Nadeen.
The remaining three — Jim, Austin, Sue — are all blond, a valid family too. So Jim's siblings are Austin and Sue.
You'll see this again: when a group must split into valid sub-groups, finding one forced sub-group (the three black-haired kids) automatically determines the other — you rarely have to test every option.
Translate the words into distance: "friends" are 1 link from Sarah, "friends of friends" are 2 links — so she invites everyone within 2 links.
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Hint 2 of 2
It's faster to count who's NOT invited: the dots that are 3+ links away or not connected to Sarah at all.
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Approach: measure each dot's link-distance from Sarah, then count the far ones
Rephrase the rule as distance. Sarah's friends are 1 segment away; their friends are 2 segments away. So an invitation reaches everyone at most 2 links from Sarah.
Sweep outward from Sarah in layers: layer 1 (direct friends), then layer 2 (their friends). Anyone left over — 3+ links away, or in a separate clump with no path to her — is not invited.
Reading the graph that way: 4 dots sit in disconnected clusters (no path to Sarah), and 2 more are reachable only after 3 links. Those are the uninvited ones.
4 + 2 = 6 classmates not invited.
You'll see this again: "friends within k steps" is a graph-distance question — spreading outward layer by layer (1 link, 2 links, 3 links…) is exactly how you find everything reachable within a fixed number of hops.
The column and the row overlap in exactly ONE square. If you just add 23 + 12, that corner square got counted twice. So the six digits sum to 23 + 12 − (the shared corner) — now you only need to find that one corner digit.
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Hint 2 of 2
Pin the corner by squeezing it from both sides. The 3-square column sums to 23, which is huge for distinct digits, so it must use the biggest ones. The 4-square row only sums to 12, so its other digits must be small. The corner has to satisfy both — try the possibilities.
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Approach: subtract the double-counted corner, then pin it
The column and row share one square, so adding both sums counts that square twice: total of all six digits = 23 + 12 − (shared corner). Everything hinges on that one corner value.
The column has 3 distinct digits summing to 23 — only {6, 8, 9} can do it (it's nearly the max 9+8+7=24). So the corner is 6, 8, or 9. The row's other 3 digits must then sum to 12 − corner: if the corner were 9 they'd need to sum to 3, and 8 → 4, both impossible for three distinct positive digits (smallest is 1+2+3=6). Only corner = 6 works, with the row finishing as 1, 2, 3.
So the six digits are {9, 8} (column-only), 6 (corner), {1, 2, 3} (row-only), and their sum is 23 + 12 − 6 = 29.
Why this transfers: whenever two overlapping groups are summed, the overlap is double-counted — subtract it once (the heart of inclusion-exclusion). And squeeze an unknown from two constraints at once instead of guessing.
Last summer 100 students attended basketball camp. Of those, 52 were boys and 48 were girls. Also, 40 students were from Jonas Middle School and 60 were from Clay Middle School. Twenty of the girls were from Jonas Middle School. How many of the boys were from Clay Middle School?
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Answer: B — 32.
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Hint 1 of 2
Every student is sorted two ways at once — boy/girl AND Jonas/Clay. When two labels overlap like that, a 2×2 grid (boys/girls across, Jonas/Clay down) keeps it all straight.
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Hint 2 of 2
You're handed the Jonas girls (20). Use a row total to fill its partner, then a column total to slide over to the answer — each blank is just 'total minus the known piece.'
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Approach: fill in a two-way table
Set up the grid. The girls' row totals 48, and 20 are from Jonas, so girls from Clay = 48 − 20 = 28.
The Clay column totals 60. Take out the 28 Clay girls: boys from Clay = 60 − 28 = 32.
Why the grid beats juggling sentences: every row and every column must hit its total, so each filled box instantly unlocks the next. Cross-check: Jonas boys = 52 − 32 = 20, and Jonas total 20 girls + 20 boys = 40 ✓ — the whole table balances.
Like a mini-Sudoku: every row AND every column holds 1, 2, 3 once each. Don't guess — hunt for a cell whose value is forced because only one number is left for it.
Still stuck? Show hint 2 →
Hint 2 of 2
Chain the forcing. Once a cell is forced, it eliminates options elsewhere, which forces the next cell. Start where you already have two of the three numbers in a line.
Show solution
Approach: forced-cell deduction, one square at a time
Middle column already shows a 2. The top row has a 1, so the top-middle cell can't be 1; being in the column-with-2 it can't be 2 either — it's forced to 3. Then the top-right cell (top row, only 2 left) is 2.
Now the right column has a 2 up top, so A (middle-right) is 1 or 3; but A's row already has the 2, and we'll see its column needs a 1: the middle row reads (left, 2, A), and the only spot left for 1 in that row makes A = 1. The right column then needs its last number, 3, for B at the bottom: B = 3.
So A + B = 1 + 3 = 4.
Why this transfers: Latin-square / Sudoku logic is never trial-and-error if you look for the cell with the fewest choices. Each forced fill shrinks the puzzle — keep chasing the most-constrained square and the grid solves itself.
Which number in the array below is both the largest in its column and the smallest in its row? (Columns go up and down, rows go right and left.)
10643211714108834591341512182593
Show answer
Answer: C — 7.
Show hints
Hint 1 of 3
Two conditions must hold at once, but they aren't equally cheap to check. One of them — being the SMALLEST in its row — only needs you to glance along a single row of five numbers. Which condition would you screen with first to throw out the most candidates fast?
Still stuck? Show hint 2 →
Hint 2 of 3
This is a search with two filters. Pick the filter that's quick to apply, run it to make a short list, then test only the survivors against the slower filter. Here: find each row's smallest, then check just those few against their column.
Still stuck? Show hint 3 →
Hint 3 of 3
There are 5 rows, so only 5 numbers can possibly be "smallest in its row." Find those 5, then check each one against its column for "largest in its column."
Show solution
Approach: screen with the cheap filter first, then test the survivors
Two conditions must both hold, so use the faster one to shrink the search. Scan each row for its smallest entry — that gives just 5 candidates instead of all 25.
Row by row the smallest entries are: 2 (top row), 7 (second row: 11, 7, 14, 10, 8), 3, 1, and 2.
Now test each candidate against its column. The 7 sits in the second column (6, 7, 3, 4, 2), where it is the largest. The others fail the column test.
So the number that is both smallest in its row and largest in its column is 7.
Why this transfers: when something must satisfy two conditions, don't check both for every item — apply the condition that's quickest to evaluate first, then test only what survives. This "cheap filter first" idea saves work in counting and search problems everywhere.
Worth knowing: a value that is the min of its row and the max of its column is called a saddle point — and there can be at most one such value in any grid.
There are 120 seats in a row. What is the fewest number of seats that must be occupied so the next person to be seated must sit next to someone?
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Answer: B — 40.
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Hint 1 of 2
Flip the goal around: 'the next person must sit next to someone' just means there's NO empty seat with both neighbors also empty. So you need every gap of empties to be small — what's the most empty seats you can leave in a row before a safe spot opens up?
Still stuck? Show hint 2 →
Hint 2 of 2
Think of each person as 'guarding' a block: themselves plus the seat on each side. To cover 120 seats with the fewest guards, give each person a full block of 3 and tile the row as (empty, person, empty) repeating.
Show solution
Approach: each person covers a block of 3 (the seated-spacing / covering idea)
Reframe: a newcomer is forced next to someone exactly when no seat is left with *both* neighbors empty. So we must break up the empties so that no two empty seats sit side by side… actually, more precisely, each occupied person can 'block' at most the seat on each side of them.
Picture repeating the pattern empty–person–empty, i.e. blocks of 3. One person per block of 3 means every empty seat is right next to an occupied one. That uses 120 ÷ 3 = 40 people, and it works.
Could 39 do it? 39 people can guard at most 39×3 = 117 seats, leaving a gap — somewhere two empties sit together and a newcomer could avoid everyone. So 40 is both enough and necessary: 40.
*Why this transfers:* 'fewest needed so every spot is covered' problems divide the total by how much one item covers (here 3) — and you confirm the floor by checking one fewer leaves a hole.
Abby, Bret, Carl, and Dana are seated in a row of four seats numbered #1 to #4. Joe looks at them and says:
"Bret is next to Carl."
"Abby is between Bret and Carl."
However each one of Joe's statements is false. Bret is actually sitting in seat #3. Who is sitting in seat #2?
Show answer
Answer: D — Dana.
Show hints
Hint 1 of 3
The clues are FALSE — that's the gift. Each false 'X is here' tells you X is NOT there, which is often more powerful than a true clue.
Still stuck? Show hint 2 →
Hint 2 of 3
Start from the fact you're handed: Bret is in #3. 'Bret is next to Carl' is false, so Carl can't be in either seat touching #3.
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Hint 3 of 3
Once Carl is forced, the second false clue ('Abby between Bret and Carl') eliminates exactly one seat for Abby — the seat strictly between Carl and Bret.
Show solution
Approach: negate each false clue, then place seats one at a time
Bret sits in #3. Clue 1 ('Bret next to Carl') is false, so Carl is NOT in #2 or #4 — the only seat left for Carl is #1.
With Carl in #1 and Bret in #3, the seat strictly between them is #2. Clue 2 ('Abby between Bret and Carl') is false, so Abby is NOT in #2. The remaining seats for Abby and Dana are #2 and #4, so Abby takes #4 and Dana takes #2.
Seat #2 holds Dana.
Why this transfers: a false 'A is next to B' statement is a constraint in disguise — flip it to 'A is NOT next to B' and use elimination just like a true clue.
Five runners finished a race: Luke, Melina, Nico, Olympia, and Pedro. Nico finished 11 minutes behind Pedro. Olympia finished 2 minutes ahead of Melina but 3 minutes behind Pedro. Olympia finished 6 minutes ahead of Luke. Which runner finished fourth?
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Answer: A — Luke.
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Hint 1 of 2
The clues compare runners to each other in a tangle. Pick one person as the zero mark and pin everybody's finish time to that single reference — which runner is mentioned most often?
Still stuck? Show hint 2 →
Hint 2 of 2
Pedro is the natural anchor (he appears in two clues). Put 0 at Pedro and write each other runner as ‘so many minutes behind Pedro,’ then just read the list in order.
Show solution
Approach: anchor everyone to one runner, then read the order
Pin everyone to Pedro at 0 (minutes behind him). Olympia is 3 behind → +3. Melina is 2 behind Olympia → +5. Luke is 6 behind Olympia → +9. Nico is 11 behind Pedro → +11.
Sorted front to back: Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11). Fourth across the line is Luke.
Why this transfers: for any ‘A is ahead/behind B’ ordering puzzle, convert every relative clue into a position on one number line anchored to a single person — the ranking then just falls out by sorting.
Don't reconstruct who-beat-whom. Each round, the 4 players split into 2 matches — so every round produces exactly 2 wins. That fixed total is your lever.
Still stuck? Show hint 2 →
Hint 2 of 2
Read the table by column instead of by row. Each column (round) must sum to 2 across all four players, so Tiyo's entry is just 2 minus the other three.
Show solution
Approach: each round's wins sum to 2
The shortcut: you don't need the matchups. Every round, four players pair into two games, so exactly 2 wins (and 2 losses) get handed out — meaning every column of the table sums to 2.
So add Lola + Lolo + Tiya down each round: 2, 2, 2, 1, 2, 1.
Tiyo fills the gap to 2 each round: 0, 0, 0, 1, 0, 1 = 000101. This transfers: when each ‘round’ has a fixed total, an unknown row is just total minus the known rows — column thinking beats casework.
Aaron, Darren, Karen, Maren, and Sharon rode on a small train that has five cars that seat one person each. Maren sat in the last car. Aaron sat directly behind Sharon. Darren sat in one of the cars in front of Aaron. At least one person sat between Karen and Darren. Who sat in the middle car?
Show answer
Answer: A — Aaron.
Show hints
Hint 1 of 2
Lock down the rigid clue first: “Aaron directly behind Sharon” glues them into a single block (Sharon then Aaron). A block of 2 has very few places it can sit, so that's your lever.
Still stuck? Show hint 2 →
Hint 2 of 2
Slide the Sharon–Aaron block across the cars and keep only the spot that still leaves room for Darren ahead of Aaron and Karen at least one car from Darren.
Show solution
Approach: glue the rigid pair into a block, then slide it
Start with the strictest clue: “directly behind” glues Sharon–Aaron into one block (S then A). Maren is fixed in car 5. A glued block of 2 only fits a handful of ways — test each.
S, A in cars 1, 2: Darren needs a car ahead of car 2, but car 1 is taken. Fail.
S, A in cars 3, 4: Darren and Karen must fill cars 1 and 2 — adjacent, with nobody between them. Breaks the spacing rule. Fail.
S, A in cars 2, 3: Darren in car 1, Karen in car 4 — three cars apart, fine. The row is Darren, Sharon, Aaron, Karen, Maren.
Only that arrangement survives, so Aaron is in the middle car.
You'll see this again as: in seating/ordering logic, attack the most restrictive clue first (a fixed seat, or two people who must be adjacent). It chops the possibilities fastest, so you test a few cases instead of all 120 orderings.
Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget and Cassie don't show theirs to anyone. Cassie says, 'I didn't get the lowest score in our class,' and Bridget adds, 'I didn't get the highest score.' What is the ranking of the three girls from highest to lowest score?
Show answer
Answer: D — Cassie, Hannah, Bridget.
Show hints
Hint 1 of 2
The hidden fact is that Cassie and Bridget have seen exactly one score besides their own: Hannah's. For either to be certain of her claim, the comparison with Hannah must settle it — so each statement secretly compares that girl to Hannah.
Still stuck? Show hint 2 →
Hint 2 of 2
In these "what can they deduce" puzzles, ask what information each speaker actually has. A confident claim is only justified if their limited view forces it — that turns each sentence into one inequality.
Show solution
Approach: convert each confident statement into a comparison with Hannah
Both girls saw only Hannah's score. Cassie says "I'm not lowest" with certainty — she can only be sure if she beat the one score she saw, so Cassie > Hannah.
Bridget says "I'm not highest" with certainty — she can only be sure if she scored below Hannah, so Hannah > Bridget.
Why this transfers: the trick is always "a certain statement reveals what the speaker can see" — the knowledge each person has is the real data, not just the words.
Students guess that Norb's age is 24, 28, 30, 32, 36, 38, 41, 44, 47, and 49. Norb says, "At least half of you guessed too low, two of you are off by one, and my age is a prime number." How old is Norb?
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Answer: C — 37.
Show hints
Hint 1 of 2
Turn each sentence into a numeric filter, then apply them one at a time — the first clue alone slices the list nearly in half.
Still stuck? Show hint 2 →
Hint 2 of 2
‘At least half of 10 guessed too low’ means at least 5 guesses are below his age, so his age beats the 5th-smallest guess. ‘Two are off by one’ means his age has a guess just below and just above it — it's wedged inside a pair of guesses differing by 2.
Show solution
Approach: translate each clue into a constraint and filter in order
Sort: 24, 28, 30, 32, 36, 38, 41, 44, 47, 49. ‘At least half too low’ needs ≥5 guesses below his age, so age > the 5th value, 36.
‘Two off by one’ means a guess sits 1 below and 1 above his age — so his age is the middle of a pair 2 apart. Above 36 the only such pairs are (36, 38) giving 37, and (47, 49) giving 48.
‘Prime’ decides it: 37 is prime, 48 is not.
Norb is 37.
Worth keeping: logic puzzles are just translation — rewrite every clue as ‘age > 36,’ ‘age between two close guesses,’ ‘age prime,’ and intersect the survivors.
Using only pennies, nickels, dimes, and quarters, what is the smallest number of coins Freddie would need so he could pay any amount of money less than a dollar?
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Answer: B — 10 coins.
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Hint 1 of 2
You only need enough coins to fill the gaps each bigger coin can't reach by itself.
Still stuck? Show hint 2 →
Hint 2 of 2
Pennies cover the 1¢–4¢ gaps below a nickel; nickels cover the gap below a dime; dimes-and-nickels cover the gap below a quarter — carry just enough of each to bridge to the next coin.
Show solution
Approach: carry just enough small coins to bridge to the next-bigger coin
To make any 1¢–4¢ ending you need 4 pennies. To reach 5¢–9¢ you need a nickel. To reach 10¢–24¢ you need a dime and a second nickel (a dime alone can't make 15¢).
Now the four pennies, two nickels, and one dime can make every amount up to 24¢. Three quarters then stack on top to push that all the way to 99¢.
Total: 4 + 2 + 1 + 3 = 10 coins.
Why this transfers: ‘cover every amount up to N’ problems are built bottom-up — carry the smallest coins/units needed to bridge each gap, then the next size stacks on top. The dime trap (it can't make 15¢ alone, so you still need a spare nickel) is the kind of gap you must check at every level.
Tiles I, II, III and IV are translated so one tile coincides with each of the rectangles A, B, C and D. In the final arrangement, the two numbers on any side common to two adjacent tiles must be the same. Which of the tiles is translated to Rectangle C?
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Answer: D — Tile IV.
Show hints
Hint 1 of 2
A number that appears on only one tile-edge has no partner to match against — so that edge can't be an interior seam. It must face outward, which pins the tile to the boundary.
Still stuck? Show hint 2 →
Hint 2 of 2
Solve constraint puzzles from the unique pieces in: anchor the forced tile first, then propagate by matching the shared-edge numbers to its neighbors, one link at a time.
Show solution
Approach: anchor on outside-only numbers, then propagate matches
Some edge numbers appear on only one tile (no other tile carries that number), so those edges must lie on the outer boundary of the 2 × 2 arrangement. Pinning those tiles into their forced corners removes most of the freedom.
From the anchored tile, walk to its neighbors by matching the shared edge number. The chain forces tile IV into rectangle C.
The numbers −2, 4, 6, 9 and 12 are rearranged according to these rules: The largest isn't first, but it is in one of the first three places. The smallest isn't last, but it is in one of the last three places. The median isn't first or last. What is the average of the first and last numbers?
Show answer
Answer: C — 6.5.
Show hints
Hint 1 of 2
The question only asks about the endpoints, so don't solve the full ordering — just figure out which numbers are banned from the ends. Each rule kicks one specific number off at least one end.
Still stuck? Show hint 2 →
Hint 2 of 2
The strategy is answer only what's asked via elimination: rather than placing all five numbers, rule out who can't be on the ends. Whoever's left must be on the ends — and you never needed the middle three.
Show solution
Approach: rule out the ends, ignore the middle
Identify the three special values: largest 12, smallest −2, median 6. Now read the rules as bans on the endpoints: the largest 'isn't first' and lives in the first three (so not last either) → 12 is off both ends; the smallest 'isn't last' and lives in the last three (so not first) → −2 is off both ends; the median 'isn't first or last' → 6 is off both ends.
Three of the five numbers are forbidden from both endpoints, so the two endpoints must be the survivors: 4 and 9.
Average of the ends: (4 + 9) ÷ 2 = 6.5.
The transferable lesson: when a puzzle asks for one feature, attack that feature directly. We never determined whether 4 or 9 is first — and we never had to, because their average is the same either way.
Amy, Bill and Celine are friends with different ages. Exactly one of the following statements is true. I. Bill is the oldest. II. Amy is not the oldest. III. Celine is not the youngest. Rank the friends from the oldest to the youngest.
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Answer: E — Amy, Celine, Bill.
Show hints
Hint 1 of 3
The hidden lever is 'exactly one is true', which means two are false. Hunt for statements that drag each other along: if assuming one true forces a second to be true, that pair can't be the single true one — so that assumption is dead.
Still stuck? Show hint 2 →
Hint 2 of 3
This is the exactly-one-true elimination: test each statement as 'the true one' and reject any that would make a second statement true as well. Watch for one claim logically implying another — that's what cracks it.
Still stuck? Show hint 3 →
Hint 3 of 3
Once a statement is known false, flip it to its opposite — a false 'Amy is not oldest' means Amy is oldest. The two falsehoods, read backwards, pin the whole order.
Show solution
Approach: test each statement as the lone truth
Try I (Bill oldest) as the true one: but Bill being oldest also makes II (Amy not oldest) true. Two trues — forbidden ⇒ I is false.
Try II (Amy not oldest) as the true one: with Bill already not oldest, the oldest would have to be Celine — but that makes III (Celine not youngest) true as well. Two trues again ⇒ II is false.
By elimination III is the lone truth, so I and II are false. Negate them: I false ⇒ Bill is not oldest; II false ⇒ Amy is the oldest. III true ⇒ Celine is not the youngest, so Celine is in the middle and Bill is youngest.
Order oldest→youngest: Amy, Celine, Bill.
The reusable idea: in 'exactly one true' puzzles, look for statements that force another to be true — those can never be the unique truth, and ruling them out usually leaves a single survivor.
Another way — check the answer choices directly:
For each ordering, count how many of I, II, III come out true; keep the one with exactly one true.
Amy, Celine, Bill: I (Bill oldest)? No. II (Amy not oldest)? No. III (Celine not youngest)? Yes — exactly one true.
Every other listed order makes zero or two statements true, so the answer is Amy, Celine, Bill.
The plus shape overlaps at the center square — that one number belongs to BOTH the row and the column. So if you add the row-sum and the column-sum together, the center gets counted twice while everyone else is counted once.
Still stuck? Show hint 2 →
Hint 2 of 2
Equal sums means each line is exactly half of (row + column). To make that half as big as possible, you want the doubled number — the center — to be the largest. Name this the 'shared-cell counts twice' idea.
Show solution
Approach: the overlap cell is counted twice — load it to maximize
Add the row total and the column total. Every number appears once except the center, which sits in both lines, so the grand total is (1+4+7+10+13) + center = 35 + center.
The two lines are equal, so each equals half of 35 + center. To push that as high as possible, put the biggest number, 13, in the center: (35 + 13) ÷ 2 = 48 ÷ 2 = 24.
Worth keeping: whenever a cell is shared by two groups whose sums you're combining, it gets double-counted — and that's a lever. Want the largest equal sums? Maximize the shared cell. (Sanity check: 24 must be achievable — center 13, then pair the rest as 1+10 and 4+7, each making 11, and 11+13 = 24. ✓)
Harry has 3 sisters and 5 brothers. His sister Harriet has S sisters and B brothers. What is the product of S and B?
Show answer
Answer: C — 12.
Show hints
Hint 1 of 2
The trap is counting from Harry's view and copying his numbers. Step back and count the whole family — total girls and total boys — then look at it fresh through Harriet's eyes.
Still stuck? Show hint 2 →
Hint 2 of 2
Nobody counts themselves as their own sibling. Harry is a boy (so he's one of the brothers from a sister's view); Harriet is a girl (so she's one of the sisters from a brother's view).
Show solution
Approach: tally the whole family, then re-count from Harriet
Build the family roster from Harry's view. Harry is a boy with 3 sisters and 5 brothers — but Harry himself is also a boy, so the family has 3 girls and 5 + 1 = 6 boys.
Now stand in Harriet's shoes. She's one of the 3 girls, so her sisters are the other 2 (she doesn't count herself): S = 2. All 6 boys are her brothers: B = 6.
So S × B = 2 × 6 = 12.
Why this transfers: in any sibling puzzle, count the family TOTAL once, then subtract the person you're asking about from their own same-gender group. The 'don't count yourself' slip is the whole point of these problems.
Logic & Word ProblemsNumber Theorywork-backwardpattern-recognition
Two players take turns removing \(1\), \(2\), \(3\), or \(4\) counters from a pile that starts with \(27\) counters. The player who takes the very last counter wins. Should you go first or second, and what is your winning plan? (Give the number of counters to take on your first move.)
Show answer
Answer: Go first and take 2 (leaving 25)
Show hints
Hint 1 of 4
There are lots of ways to start, but only one way to finish (grab the last few counters). When the END is clearer than the start, work backward from the end.
Still stuck? Show hint 2 →
Hint 2 of 4
Ask: how many counters can I leave for my opponent so that no matter what they take (\(1\), \(2\), \(3\), or \(4\)), I can take the rest and grab the last one?
Still stuck? Show hint 3 →
Hint 3 of 4
If you leave exactly \(5\), you win: whatever they take, you take the rest of those \(5\). Now work back one more step. To be able to leave \(5\) next time, what should you leave now?
Show solution
Approach: Working backward from the finish to find the safe positions
To win you want to take the last counter, so after your final move there are \(0\) left. Work backward from there.
The magic numbers to leave for your opponent are the multiples of \(5\). Leaving \(5\) works: if they take \(k\) (one of \(1, 2, 3, 4\)), you take the remaining \(5 - k\) and grab the last counter.
So every turn you hand them a multiple of \(5\): \(25, 20, 15, 10, 5, 0\). Whatever they remove, you remove enough to land on the next multiple of \(5\).
Since \(27 = 25 + 2\), you go FIRST and take \(2\), leaving \(25\). After that always leave a multiple of \(5\); your last move leaves \(0\), so you took the last counter and win.
Geometry & MeasurementLogic & Word Problemsaccount-for-all-possibilitiesconsider-extreme-cases
A cube has six faces (flat surfaces). Now cut a cube in half. How many faces does half a cube have? Try to think of all the different ways the cube might be cut in half.
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Answer: No single answer: 6 for a flat straight cut, 5 for a slanted cut, more for a staircase cut — the point is to define what 'half' means
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Hint 1 of 4
There is no single 'right' answer here! The number of faces depends on HOW you slice. Start by picturing the simplest cut you can imagine.
Still stuck? Show hint 2 →
Hint 2 of 4
Try a straight cut straight down through the middle, like slicing a block of cheese. You get a smaller box. Count its flat surfaces carefully — don't forget the brand-new face the knife made.
Still stuck? Show hint 3 →
Hint 3 of 4
Now try a slanted cut, going corner to corner. Count again. Did you get a different number? What does that tell you?
Show solution
Approach: Open-ended exploration: the answer depends on the cut
This problem is meant to be open-ended; its value is in the discussion, not in one magic number.
Straight cut down the middle: you get a smaller rectangular box, which has 6 flat faces — so this answer is 6 (one is the new face the knife made).
Slanted (corner-to-corner) cut: the piece is a wedge or triangular prism, and you count 5 faces. (Some people blurt out 3 before noticing the original faces are still partly there.)
Staircase cut: if the cut is a jagged staircase that still splits the cube into two equal-volume pieces, every step adds faces, so the count can be as big as you like.
The real lesson: 'half a cube' is ambiguous. Once you notice that, many answers become reasonable, and you start asking better questions: must the pieces match exactly? must the cut be flat?
Arithmetic & OperationsLogic & Word Problemsidentifying-relevant-datatranslate-text-into-mathematics
A class from a town of 4,300 people takes a trip to a mountain 120 km away. The class has 500 dollars in its treasury. The whole trip cost 360 dollars. That 360 dollars paid for the bus (110 dollars) plus a rope-walk activity that costs the same amount for each of the 25 students. (a) How much did the rope-walk cost for one student? (b) Which numbers in the problem were NOT needed to answer part (a)?
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Answer: 10 dollars per student; population, distance, and treasury are not needed
Show hints
Hint 1 of 3
Before you compute, sort the numbers into two piles: ones that change the rope-walk price per student, and ones that are just background story.
Still stuck? Show hint 2 →
Hint 2 of 3
The 360 dollars total is made of two things: the one bus fee (110 dollars) plus the rope-walk paid once for each of the 25 students.
Still stuck? Show hint 3 →
Hint 3 of 3
Subtract the bus fee from the total to get just the rope-walk money. Then split that evenly among the 25 students.
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Approach: Strip the irrelevant data, then subtract and divide
The 360 dollars is the bus fee plus all the rope-walk fees. Take out the bus fee first: \(360 - 110 = 250\) dollars.
That 250 dollars is shared equally by 25 students, so each student's rope-walk costs \(250 \div 25 = 10\) dollars.
(a) The rope-walk costs 10 dollars per student.
(b) The town's population (4,300), the distance (120 km), and the 500-dollar treasury are never used — they are background story, not needed for part (a).
Logic & Word ProblemsCounting & Probabilityconsider-extreme-casesaccount-for-all-possibilitiespattern-recognition
In a fairy tale, a hero must collect golden hairs from a devil's head. The devil's grandmother is blind, so she pulls hairs out at random. The devil has \(5\) hairs in all, and \(3\) of them are golden. How many hairs must the blind grandmother pull to be sure she has at least one golden hair? Then think bigger: the devil has \(x\) hairs, \(y\) of which are golden. How many hairs must be pulled to be sure of getting at least \(z\) golden hairs? (Here \(z \le y \le x\), and all are whole numbers.)
Show answer
Answer: 3 hairs in the small case; (x−y)+z hairs in general
Show hints
Hint 1 of 4
The word 'sure' is the key. You can't count on getting lucky. Ask yourself: what is the unluckiest possible order in which the hairs could come out?
Still stuck? Show hint 2 →
Hint 2 of 4
Pretend the grandmother has the worst luck in the world: she keeps grabbing only the non-golden hairs first. With \(5\) hairs and \(3\) golden, how many hairs are NOT golden?
Still stuck? Show hint 3 →
Hint 3 of 4
Once every non-golden hair is gone, the very next pull MUST be golden. There are \(5-3=2\) non-golden hairs, so after pulling those \(2\), the next pull (the 3rd) is guaranteed golden.
Show solution
Approach: Worst-case (pigeonhole-style) reasoning
'Sure' means it must work even with the worst luck: every non-golden hair comes out first. There are \(5-3=2\) non-golden hairs.
So in the worst case the first 2 pulls are both non-golden; then only golden hairs are left, so the 3rd pull must be golden. She must pull \(2+1=3\) hairs.
General case: there are \(x-y\) non-golden hairs. In the worst case she pulls all \(x-y\) of them first and gets no gold; after that every remaining hair is golden.
To collect \(z\) golden hairs she pulls \((x-y)+z\) hairs. Check with \(x=5, y=3, z=1\): \((5-3)+1=3\), which matches. (This is the same idea behind pigeonhole problems.)
Logic & Word ProblemsNumber Theorywork-backwardpattern-recognition
Same game as before: take \(1\), \(2\), \(3\), or \(4\) counters from a pile of \(27\). But now the player who takes the LAST counter LOSES. What is your winning plan? (Give the number of counters to take on your first move.)
Show answer
Answer: Go first and take 1 (leaving 26)
Show hints
Hint 1 of 4
Now losing means being forced to take the final counter. So you want to hand your opponent exactly \(1\) counter at the very end — then they must take it and lose.
Still stuck? Show hint 2 →
Hint 2 of 4
Work backward from leaving \(1\). What is the next safe number to leave below that?
Still stuck? Show hint 3 →
Hint 3 of 4
Just like the last game used multiples of \(5\), this game uses numbers that are \(1\) more than a multiple of \(5\): \(1, 6, 11, 16, 21, 26\).
Show solution
Approach: Working backward to force the opponent to take the last counter
You win by forcing your opponent to take the last counter, so on your final turn you want to leave exactly \(1\).
Leaving \(1\) is safe (they must take it and lose). Working backward, the safe numbers to leave are one more than a multiple of \(5\): \(1, 6, 11, 16, 21, 26\).
From any of these, whatever your opponent takes (\(k\)), you take \(5 - k\) to get back to the next safe number.
Since \(27 = 26 + 1\), you go FIRST and take \(1\), leaving \(26\). Then always leave a number that is \(1\) more than a multiple of \(5\); your last move leaves \(1\), your opponent must take it, and they lose.
Counting & ProbabilityLogic & Word Problemsconsidering-extreme-casesaccounting-for-all-possibilitieslogical-reasoning
A drawer has 7 blue socks and 7 red socks, all jumbled together. You reach in (in the dark) and pull out socks. (1) How many socks must you grab to be CERTAIN of getting a matching pair of some color? (2) Now a harder, different question: how many must you grab to be CERTAIN of getting two BLUE socks specifically? (Imagine the worst possible luck.)
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Answer: Any matching pair: 3 socks. Two blue socks specifically: 9 socks
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Hint 1 of 4
'A matching pair of some color' means two blues OR two reds. There are only two colors. Think about the worst case: what is the most socks you could grab and still NOT have a pair?
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Hint 2 of 4
If you grabbed 2 socks of different colors (one blue, one red), you have no pair yet. But the very next sock must match one of them!
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Hint 3 of 4
For part (2), 'two blue' is much pickier. Worst luck: you keep pulling out red socks. How many reds are in the drawer? You might pull every one of them before a blue shows up.
Show solution
Approach: Considering the worst case — pigeonhole vs. a specific color
Part 1, a matching pair of any color: with only two colors, after you grab 2 socks the unluckiest result is one blue and one red — no pair yet. But the 3rd sock has to be blue or red, so it MUST match one of the two you already hold. So 3 socks guarantee a matching pair. (You can also list the patterns of 3 socks: BBB, BBR, BRR, RRR — every one contains a pair.)
Part 2, two BLUE socks specifically: this is a pickier demand. Imagine pulling out reds again and again with terrible luck. There are 7 red socks, so you could pull all 7 reds before any blue appears. After those 7 reds you still need 2 blue socks, so in the worst case you need \(7+2=9\) socks.
The lesson: read the question carefully! 'A matching pair of any color' (3 socks) and 'two of a specific color' (9 socks) have completely different answers.
Counting & ProbabilityLogic & Word Problemspigeonholelogical-reasoning
At a party there are 6 people, and everyone knows at least one other person there. Show that at least 2 people know the exact same number of the others. (Knowing is mutual: if A knows B, then B knows A.)
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Answer: at least 2 people match
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Hint 1 of 4
Each person knows somewhere between 1 person (the smallest, since everyone knows at least one) and 5 people (everyone else).
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Hint 2 of 4
So each person's 'number of friends here' is one of the values 1, 2, 3, 4, or 5. How many choices is that?
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Hint 3 of 4
That's only 5 possible 'friend-count' labels, but there are 6 people. Make the 5 labels your boxes and drop each person into the box for their count.
Show solution
Approach: Pigeonhole — 6 people, only 5 possible friend-counts
Each of the 6 people knows at least 1 other and at most 5 others, so each person's number of acquaintances is one of \(1, 2, 3, 4, 5\).
That's only 5 possible values. Make those 5 values into 5 boxes ('knows 1', 'knows 2', ..., 'knows 5').
Put each of the 6 people into the box for how many people they know. With 6 people and only 5 boxes, some box holds at least 2 people.
Those 2 people know the same number of others, so at least \(2\) people must match.
Arithmetic & OperationsLogic & Word ProblemsRatios, Rates & Proportionsidentifying-relevant-datavisual-representationlogical-reasoning
A worker bikes to work. The trip is 3 km and he usually rides at 15 km/h. One day, after going 1 km, he gets a flat tire, so he pushes his bike the rest of the way, arriving 20 minutes late. At work he fixed the tire and rode all the way home as usual. Over the whole round trip (there and back), how many more kilometers did he ride than he walked?
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Answer: 2 km more by bike
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Hint 1 of 4
Draw the trip as a straight line: home to work, then work back home. Mark the flat (1 km from home). Shade the parts he rode and the parts he walked.
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Hint 2 of 4
Going to work: he rode the first 1 km, then walked the last 2 km. Coming home: he rode all 3 km. Now compare total riding to total walking.
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Hint 3 of 4
The 2 km from the flat to work was walked once (going) and ridden once (coming home), so those cancel. What part of the route did he ride but never walk?
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Approach: Picture the round trip and cancel the matching segments
Picture the round trip: home —(ride 1 km)— flat —(walk 2 km)— work, then work —(ride 3 km)— home.
He rode 1 km going + 3 km coming = 4 km. He walked 2 km. So he rode \(4 - 2 = 2\) km more than he walked.
A neat way to see it: the 2 km from the flat to work was walked once and ridden once, so it cancels. Only the first 1 km was ridden both directions and never walked, counting twice (\(2 \times 1 = 2\) km).
So he rode 2 km more than he walked. (Notice everything except that 1 km — even the speed and lateness — is unnecessary for this question.)
Logic & Word ProblemsGeometry & Measurementsymmetrylogical-reasoningconsidering-extreme-cases
A hunter leaves camp, walks 10 miles due north in a straight line, and stops for lunch. After lunch he again walks 10 miles due north in a straight line — and discovers he is back at camp! On a round Earth, where is the hunter's camp? ('North' always means 'toward the North Pole,' and his straight line follows the curve of the Earth like a taut string on a globe.)
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Answer: The South Pole
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Hint 1 of 3
On a globe, 'north' is not one fixed compass direction forever. It always means 'toward the North Pole.' Is there a special place where 'north' behaves strangely?
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Hint 2 of 3
Think about the South Pole. From the South Pole, which direction is north?
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Hint 3 of 3
From the South Pole, EVERY direction is north. So if the camp is at the South Pole, walking 'north' takes you out, and walking 'north' again along the same taut-string path on the far side brings you right back.
Show solution
Approach: Use the special symmetry of the pole
'North' means toward the North Pole. On a flat map that feels like one fixed direction, but on a round Earth it changes with where you stand. The trick is to find the one special spot where this matters.
Stand exactly at the South Pole. The North Pole is straight up and over the globe in every direction, so from the South Pole every direction you face is 'north.'
Put the camp at the South Pole. Walking 10 miles 'north' carries the hunter out along a great circle; walking 'north' again continues along that same circle, which loops back toward the pole on the far side and returns him to the South Pole.
Because all directions from the South Pole are identical (all 'north'), a 'north then north' walk can close on itself. No ordinary spot has that symmetry, so the camp is at the South Pole.
Logic & Word ProblemsCounting & ProbabilityNumber Theoryaccount-for-all-possibilitiesreduce-and-expandcounting-principle
How many \(2\)-digit whole numbers are there? Then generalize: how many \(n\)-digit whole numbers are there, where \(n\) is a whole number bigger than \(1\)? (Hint to start small: first try counting using only the digits \(0\) and \(1\), then using \(0,1,2,3\).)
Show answer
Answer: 90 two-digit numbers; in general 9 × 10^(n−1)
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Hint 1 of 4
There are two nice ways to do this. You can count a range of numbers, or you can count digit by digit using the multiplication (counting) principle.
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Hint 2 of 4
Range way: the smallest \(2\)-digit number is \(10\) and the biggest is \(99\). How many whole numbers are there from \(10\) to \(99\), counting both ends?
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Hint 3 of 4
Counting way: the first digit can't be \(0\) (or it wouldn't be a \(2\)-digit number), so how many choices does it have? The second digit can be anything \(0\) through \(9\). Multiply the two counts.
Show solution
Approach: Count a range, or use the multiplication counting principle
Way 1 (count a range): the whole numbers from \(a\) to \(b\) number \(b-a+1\). The 2-digit numbers go from 10 to 99, so there are \(99-10+1=90\).
Way 2 (counting principle): the first digit can be \(1,\dots,9\) (9 choices, no leading 0) and the second digit can be \(0,\dots,9\) (10 choices), giving \(9\times10=90\).
Generalizing to \(n\) digits: 9 choices for the leading digit and 10 for each of the other \(n-1\) digits, so \(9\times10^{\,n-1}\).
Check: \(n=2\) gives \(9\times10=90\); \(n=3\) gives \(9\times100=900\), matching the three-digit numbers from 100 to 999.
Logic & Word ProblemsArithmetic & Operationswork-backwardlogical-reasoning
You have only a \(5\)-liter bucket and an \(11\)-liter bucket and as much water as you want. How can you end up with exactly \(7\) liters in the big (\(11\)-liter) bucket?
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Answer: 7 liters (make 1 L, then top off the small bucket to pour off exactly 4 L)
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Hint 1 of 4
Work backward from the goal. You want \(7\) liters in the \(11\)-liter bucket. How much EMPTY space is left in that bucket when it holds \(7\)? (\(11-7\).)
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Hint 2 of 4
There are \(4\) liters of empty space. You could create exactly \(4\) empty liters by pouring from a full \(11\)-liter bucket into the small bucket — but only if the small bucket already had \(1\) liter in it (so it can take just \(4\) more).
Still stuck? Show hint 3 →
Hint 3 of 4
So now you only need to make exactly \(1\) liter. Try filling the big bucket and pouring out \(5\) liters twice: \(11-5-5=1\) liter is left over.
Show solution
Approach: Working backward, then doing the steps forward
Work backward: \(7\) liters in the big bucket leaves \(4\) liters of empty space. To pour off exactly \(4\) liters into the \(5\)-liter bucket, the small bucket must already hold \(1\) liter (so it only has room for \(4\) more). And to get exactly \(1\) liter, notice \(11-5-5=1\).
Now the forward steps. Fill the \(11\)-liter bucket. Pour into the \(5\)-liter bucket and dump it out; do this twice. After pouring out \(5+5=10\), the big bucket holds \(1\) liter.
Pour that \(1\) liter into the empty \(5\)-liter bucket. Fill the \(11\)-liter bucket again (now it holds \(11\)).
Pour from the big bucket to fill the small bucket the rest of the way. The small bucket had \(1\), so it takes \(4\) more. The big bucket now has \(11-4 = 7\) liters. Done!
Arithmetic & OperationsLogic & Word Problemslogical-reasoningwork-backwardidentifying-relevant-data
Mr. Mayer takes the 7:25 bus to his office in the morning, then walks home in the afternoon. The whole day's travel (bus there + walk back) takes 1 hour 10 minutes. If he walked BOTH ways, it would take 1 hour 50 minutes. How long would it take him to take the bus BOTH ways?
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Answer: 30 minutes by bus both ways
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Hint 1 of 4
Give names to the two one-way times: let \(W\) = time to walk one way, \(B\) = time to bus one way.
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Hint 2 of 4
Walking both ways means \(W + W = 1\) hour 50 minutes. So \(2W = 110\) minutes — find \(W\).
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Hint 3 of 4
Bus-then-walk means \(B + W = 1\) hour 10 minutes = 70 minutes. Once you know \(W\), subtract to get \(B\).
Show solution
Approach: Work backward to the one-way times, then double the bus time
Let \(W\) be the one-way walking time and \(B\) the one-way bus time, in minutes.
Walking both ways takes 1 h 50 min = 110 min, and that is \(W + W\): \(2W = 110\), so \(W = 55\) min.
Bus-then-walk takes 1 h 10 min = 70 min, and that is \(B + W\): \(B + 55 = 70\), so \(B = 15\) min.
Both ways by bus takes \(2B = 2 \times 15 = 30\) minutes. (The 7:25 start time is not needed for this; with it, he would arrive at 7:40.)
Logic & Word ProblemsCounting & ProbabilityArithmetic & Operationsaccount-for-all-possibilitiesorganizing-datareduce-and-expandpattern-recognition
A strip of \(15\) unit squares is cut into 'pieces.' Each piece is a run of \(1, 2, 3, 4,\) or \(5\) squares, and we use at most \(5\) pieces. List all the ways to write \(15\) as a sum of pieces under these rules (order of the pieces doesn't matter). Then connect to Gauss: one of the \(5\)-piece answers is \(5 + 4 + 3 + 2 + 1\), the numbers \(1\) through \(5\). Use this to find a quick formula for \(1 + 2 + 3 + \cdots + n\).
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Answer: 1 way with 3 pieces, 5 with 4 pieces, 12 with 5 pieces; and 1+2+…+n = n(n+1)/2
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Hint 1 of 4
Each piece is at most \(5\) squares and they must add to \(15\). What is the smallest number of pieces you could possibly use?
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Hint 2 of 4
Since \(5 + 5 + 5 = 15\), you can't do it in fewer than \(3\) pieces. So organize your search into \(3\)-piece, \(4\)-piece, and \(5\)-piece cases.
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Hint 3 of 4
In each case you're writing \(15\) as a sum of that many numbers, each between \(1\) and \(5\). Always list the biggest part first so you don't miss any or repeat any.
Show solution
Approach: Organized casework on number of pieces, then Gauss pairing
The total is 15, each piece is 1 to 5 squares, at most 5 pieces. Since the biggest piece is 5 and \(5\times3=15\), you need at least 3 pieces.
Gauss: in \(1+2+3+4+5\), pair \(1+5=6\), \(2+4=6\), middle 3; each pair adds to \(n+1\). For \(1+2+\cdots+n\), pairing the ends always gives \(n+1\), so \(1+2+\cdots+n=\dfrac{n(n+1)}{2}\). Check: \(1+2+3+4+5=\dfrac{5\times6}{2}=15\), and \(1+2+\cdots+100=\dfrac{100\times101}{2}=5050\).
Logic & Word ProblemsArithmetic & Operationswork-backwardintelligent-guessing-and-testing
Jim asks a classmate how old she is. She answers, 'I was 14 the day before yesterday, but I'll be 17 next year.' On what day is she talking, and when is her birthday?
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Answer: She is speaking on January 1, and her birthday is December 31
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Hint 1 of 4
This sounds impossible, but it isn't! The secret is that a few days near New Year's can touch several different calendar years at once.
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Hint 2 of 4
If she was 14 just a couple of days ago but turns 17 'next year,' her age has to climb 14, 15, 16, 17. That's a lot of birthdays squeezed close together — so her birthday must sit right at the edge of the year.
Still stuck? Show hint 3 →
Hint 3 of 4
Try guessing her birthday is December 31. Then try having the conversation on January 1. Walk through the days: the day before yesterday, yesterday, today, the next December 31, and the one after that.
Show solution
Approach: Guess a year-edge birthday, then test the timeline
This only seems like a paradox if you forget that 'next year' on the calendar can be just a day or two away. Let her birthday be December 31 and the conversation happen on January 1.
The day before yesterday = December 30: she hadn't yet had her Dec 31 birthday, so she was 14 (matches 'I was 14 the day before yesterday').
Yesterday = December 31: she turned 15. Today = January 1: she is 15.
This coming December 31 she turns 16; the December 31 after that she turns 17. Since today is January 1, that birthday lands in the NEXT calendar year (matches 'I'll be 17 next year').
Everything fits: she is speaking on January 1, and her birthday is December 31.
Arithmetic & OperationsLogic & Word Problemspattern-recognitionconsider-extreme-cases
A truck carries 4,000 crates. At its first stop it drops off half the crates. At the second stop it drops off half of what is left. At the third stop, half of the new remainder. If this pattern keeps going, at which stop will the LAST crate be dropped off?
Show answer
Answer: There is no last stop — half always remains, so the question has no answer (a peek at infinity)
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Hint 1 of 4
Don't track how many get dropped off. Track how many are still ON the truck after each stop.
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Hint 2 of 4
Start halving: 4000, then 2000, then 1000, ... What stays on the truck each time?
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Hint 3 of 4
After every single stop, exactly half of the crates remain on the truck. Ask yourself: can the truck ever hit zero this way?
Show solution
Approach: Track what stays, and realize it never reaches zero
Follow the crates left on the truck. Dropping off half means half stay: \(4000\to2000\to1000\to500\to250\to125\to\dots\)
After each stop, half of whatever is on the truck stays on the truck, so no matter how many stops happen, something is always still aboard.
(Once the number gets odd, like 125, you can't literally split it evenly — the problem ignores that on purpose, and arguing about it is part of the fun.)
So there is no last stop: the amount on the truck gets closer and closer to 0 but never reaches it. The 'success' here is realizing there is no answer — a neat doorway to thinking about infinity.
Counting & ProbabilityLogic & Word Problemscount-the-complementconsider-extreme-cases
A tennis tournament has 61 players. In any round with an odd number of players, one player gets a 'bye' (they skip that round and advance without playing). Every match is played until someone wins; the loser is knocked out. How many matches are played in all before one champion is left undefeated?
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Answer: 60 matches
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Hint 1 of 4
First make sure you understand a 'bye': it just means a player moves on to the next round without playing, because there was an odd number of players.
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Hint 2 of 4
You COULD draw the whole bracket round by round and add up the matches. That works, but it's a lot of bookkeeping.
Still stuck? Show hint 3 →
Hint 3 of 4
Try a smarter idea: instead of counting winners, count LOSERS. Every match knocks out exactly one player.
Show solution
Approach: Count losers, not winners
Bracket way (the long way): Round 1 has 61 players (one bye), 30 matches, 31 left; Round 2: 15 matches, 16 left; Round 3: 8 matches, 8 left; Round 4: 4 matches; Round 5: 2 matches; Round 6: 1 match. Total \(30+15+8+4+2+1=60\).
The clever way: the champion is the only player who never loses, and everyone loses exactly once, by being knocked out in one match.
A bye is not a match and knocks no one out. Each match knocks out exactly one player, and we must knock out everyone except the champion.
So the number of matches is \(61-1=60\). Counting losers skips all the messy bye bookkeeping.
Logic & Word ProblemsCounting & ProbabilityArithmetic & Operationsorganizing-datalogical-reasoning
Six patients A, B, C, D, E, F wait at a dentist. Their treatment times are A = 15 min, B = 30 min, C = 10 min, D = 10 min, E = 20 min, F = 5 min. The dentist wants the total waiting time of all patients added together to be as small as possible. What is that smallest possible total waiting time (in minutes)?
Show answer
Answer: 145 minutes (treat shortest first: F, C, D, A, E, B)
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Hint 1 of 4
The first patient seen waits 0 minutes. The second waits through the first one's treatment. Everyone waits through everyone seen before them.
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Hint 2 of 4
Whoever goes FIRST makes all 5 others wait through their treatment. So a long treatment early on is very costly.
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Hint 3 of 4
To keep the total small, put the shortest treatments first and the longest last.
Show solution
Approach: Shortest-job-first scheduling
The first patient's time is waited through by all 5 others, the second's by 4, then 3, 2, 1, 0. To make the total small, the biggest counts should multiply the smallest times — so treat the shortest patient first.
The times in order are 5, 10, 10, 15, 20, 30, which is F, C, D, A, E, B.
Logic & Word ProblemsArithmetic & OperationsCounting & Probabilitylogical-reasoningconsidering-extreme-cases
You win a lottery! There are three piles of bills: a 100-dollar pile, a 50-dollar pile, and a 10-dollar pile. You may take 10 bills from one pile, 5 bills from another, and 1 bill from the third (you choose which pile gets which count). Matching the counts to win the most money, how many dollars do you win?
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Answer: 1260 dollars
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Hint 1 of 4
This is like the dentist problem flipped: now you want the total to be as BIG as possible.
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Hint 2 of 4
Each pile's bill value gets multiplied by one of the counts 10, 5, or 1. Which count do you want on the most valuable bill?
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Hint 3 of 4
Put the biggest count on the biggest bill: take 10 bills of 100 dollars, 5 bills of 50 dollars, 1 bill of 10 dollars.
Show solution
Approach: Pair the largest multiplier with the largest value
Your winnings are (some count) times each pile's value, added up. To make that largest, give the largest count to the largest bill.
Take 10 bills from the 100-dollar pile, 5 from the 50-dollar pile, and 1 from the 10-dollar pile.
Geometry & MeasurementLogic & Word Problemsconsidering-extreme-casesaccount-for-all-possibilities
On the round Earth, from how many starting points can you walk exactly \(1\) mile south, then \(1\) mile east, then \(1\) mile north and end up right where you started? Describe ALL such points.
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Answer: Infinitely many: the North Pole plus infinitely many circles near the South Pole
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Hint 1 of 4
Look at the extreme spots on Earth: the poles. Try starting at the North Pole and check what happens.
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Hint 2 of 4
There is more than one answer! Think about the South Pole area. The 'east' walk goes around a circle of latitude. When would walking \(1\) mile east bring you back to where you started that leg?
Still stuck? Show hint 3 →
Hint 3 of 4
If your \(1\) mile south lands you on a tiny circle whose distance all the way around is exactly \(1\) mile, then walking \(1\) mile east loops you all the way around, back to the same spot. Then \(1\) mile north returns you to start.
Show solution
Approach: Considering extreme cases (the poles) and accounting for all possibilities
North Pole: start there, walk \(1\) mile south, then \(1\) mile east along a latitude circle, then \(1\) mile north — back at the pole. It works.
Near the South Pole: find the circle that is exactly \(1\) mile all the way around. Stand anywhere \(1\) mile NORTH of it. Walking south lands you on the little circle, walking \(1\) mile east loops you all the way around back to the same point, and walking north returns you to start. Every point on that bigger circle works — infinitely many.
More possibilities: the little circle could instead be \(\frac{1}{2}\) mile around (the east walk loops it twice), or \(\frac{1}{3}\) mile around (three times), and so on for any whole number of loops.
Full answer: the North Pole, plus — for every whole number \(n = 1, 2, 3, \dots\) — every point \(1\) mile north of the southern circle whose distance around is \(\frac{1}{n}\) mile. That is infinitely many starting points.
Logic & Word ProblemsGeometry & Measurementproof-by-contradictionparity
Is it possible to draw a single straight line that crosses every one of the \(999\) sides of a \(999\)-sided polygon? Explain why or why not.
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Answer: No — it is impossible (a parity argument, since 999 is odd)
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Hint 1 of 4
Suppose it IS possible, and look for something that goes wrong (this is proof by contradiction). A straight line splits the plane into two sides — call them 'left' and 'right'.
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Hint 2 of 4
Walk around the polygon vertex by vertex. Every time you cross the line, you switch from one side to the other. So crossing a side means the two endpoints of that side are on opposite sides of the line.
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Hint 3 of 4
If the line crosses ALL 999 sides, then every pair of neighboring vertices is on opposite sides — the vertices must perfectly alternate left, right, left, right... all the way around.
Show solution
Approach: Proof by contradiction using parity
Suppose, for contradiction, that one straight line crosses all 999 sides. The line cuts the plane into two halves.
Walk around the polygon from vertex to vertex. Crossing a side means stepping over the line, switching half-planes, so a crossed side has its two endpoints on opposite sides.
If every one of the 999 sides is crossed, the vertices must alternate left, right, left, right around the whole loop and return to the start.
Returning to the start after alternating only works with an even number of vertices. Since \(999\) is odd, perfect alternation around the loop is impossible — a contradiction. So no such line exists.
Counting & ProbabilityLogic & Word Problemsasking-key-questionspigeonholeconsidering-extreme-cases
A drawer has \(7\) pairs of blue socks and \(7\) pairs of red socks, all jumbled together. Reaching in the dark, how many socks must you grab at once to be SURE you have a matching pair (two of the same color)?
Show answer
Answer: 3 socks
Show hints
Hint 1 of 3
The numbers \(7\) and \(7\) are a distraction. Ask the key question: assuming the worst possible luck, how many socks could you pull out and STILL not have a matching pair?
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Hint 2 of 3
There are only two colors. The most you could grab without a match is one blue and one red — just \(2\) socks.
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Hint 3 of 3
Now think about the very next sock. What color can it possibly be?
Show solution
Approach: Asking the key question — the pigeonhole worst case
Ask the key question: with the worst luck, how many socks can you draw and still have no match?
There are only two colors. The unluckiest grab gives you one blue and one red — \(2\) socks, no match yet.
But the third sock you pull MUST be blue or red, and either way it matches one you already hold. So \(3\) socks are always enough (and \(2\) is not). The numbers \(7\) and \(7\) never matter — only that there are \(2\) colors.
Geometry & MeasurementCounting & ProbabilityLogic & Word Problemsasking-key-questionsseeking-complementssymmetry
Ten checkers are set up in a triangle pointing UP, with rows of \(1, 2, 3,\) and \(4\). What is the fewest checkers you must move so the triangle points DOWN instead?
Show answer
Answer: 3 checkers
Show hints
Hint 1 of 3
Instead of asking which checkers to move, ask the opposite (complement) question: which checkers can STAY where they are because they're already in the right spot for the downward triangle too?
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Hint 2 of 3
Lay the downward triangle on top of the upward one (same overall outline, flipped). Many checkers overlap — those don't need to move.
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Hint 3 of 3
Count how many checkers sit in the overlap and can stay. The answer is \(10\) minus that number.
Show solution
Approach: Seeking complements — count the checkers that can stay
Ask the complement question: how many checkers can stay put? A checker can stay if its spot is part of BOTH the up-triangle and the down-triangle.
When you overlay the upward triangle and its upside-down version, \(7\) of the \(10\) checkers land on shared spots and don't have to move. Only \(3\) checkers are out of place — the single checker at the top point and the two checkers at the bottom corners.
Counting & ProbabilityLogic & Word Problemsseeking-complementsasking-key-questions
A tournament has \(36\) players. One loss knocks you out. How many games must be played to crown a single champion? Then: how would the answer change if it took TWO losses to be knocked out?
Show answer
Answer: 35 games (single elimination); 70 or 71 games if two losses are needed
Show hints
Hint 1 of 3
Don't count games from the winner's side — that's hard. Ask the complement question: how many LOSERS are there, and how does a loss relate to a game?
Still stuck? Show hint 2 →
Hint 2 of 3
Each game produces exactly one loser. So the number of games equals the total number of losses handed out.
Still stuck? Show hint 3 →
Hint 3 of 3
Everyone except the one champion gets eliminated. For one-loss: \(35\) players must be eliminated, so \(35\) losses, so \(35\) games. For two-loss: each eliminated player needs \(2\) losses.
Show solution
Approach: Seeking complements — count losses, not wins
Ask the complement question: count losses, not wins. Every game makes exactly one loser, so the number of games equals the number of losses.
One loss to eliminate: out of \(36\) players, exactly \(1\) becomes champion and the other \(35\) are each eliminated by \(1\) loss. That is \(35\) losses, so \(35\) games.
Two losses to eliminate: each of the \(35\) eliminated players must collect \(2\) losses, giving \(35 \times 2 = 70\) losses. The champion might also pick up a loss along the way (one is allowed), so the total is \(70\) games if the champion never lost, or \(71\) if the champion lost exactly once before winning.
Number TheoryLogic & Word Problemslogical-reasoningpattern-recognition
In a leap year, January has \(31\) days, February \(29\), March \(31\). Using the fact that a week repeats every \(7\) days, January 1 and April 1 fall on the same weekday because the number of days from January 1 to April 1 is a multiple of \(7\). How many days is that?
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Answer: 91 days (= 7 x 13)
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Hint 1 of 4
Two dates land on the same weekday when the number of days between them is a multiple of \(7\). So count the days from January 1 to April 1.
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Hint 2 of 4
The days from January 1 to April 1 equal the lengths of January, February, and March added together.
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Hint 3 of 4
In a leap year, January has \(31\) days, February has \(29\), March has \(31\). Add them up.
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Approach: Count days, then check divisibility by 7
Two dates fall on the same weekday exactly when the number of days between them divides evenly by \(7\), since the weekday pattern repeats every \(7\) days.
From January 1 to April 1 the days you pass through are all of January, February, and March: \(31 + 29 + 31 = 91\) days.
Divide by \(7\): \(91 = 7 \times 13\), an exact multiple with no remainder. So April 1 lands on the same weekday as January 1.
Bonus: April 1 to July 1 is \(30 + 31 + 30 = 91\) days too, so July 1 also matches — that is why January, April, and July all start on the same weekday in a leap year.
Counting & ProbabilityLogic & Word Problemspigeonholelogical-reasoning
In a town of 10 people, everyone has at least 1 friend (friendship is mutual). Show that at least 2 people have the same number of friends.
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Answer: at least 2 people have the same friend-count
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Hint 1 of 4
This is the party problem again. The people are the objects; their friend-count is the label.
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Hint 2 of 4
Each person has at least 1 friend and at most 9 friends (everyone else). List the possible friend-counts.
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Hint 3 of 4
The counts are \(1, 2, \dots, 9\) — that's 9 possible values (boxes) for 10 people.
Show solution
Approach: Pigeonhole — 10 people, only 9 possible friend-counts
Each of the 10 people has at least 1 friend and at most 9 friends (everyone else in town), so each person's friend-count is one of \(1, 2, 3, \dots, 9\) — 9 possible values.
Make these 9 values the boxes and put each person into the box equal to their number of friends.
With 10 people but only 9 boxes, some box holds at least 2 people.
Logic & Word ProblemsCounting & Probabilitypigeonholecounterexamplelogical-reasoning
Someone claims: 'If there are more leaves than trees, then at least 2 trees must have the same number of leaves.' Is this true? Explain.
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Answer: No — it is not always true
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Hint 1 of 4
Try the obvious pigeonhole setup, then test it on a tiny example to see if it really works.
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Hint 2 of 4
If trees are the boxes and leaves are the objects, pigeonhole says some tree has lots of leaves — but does it say two trees MATCH?
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Hint 3 of 4
Could one tree have 0 leaves? In the 'friends' problems, the trick worked because everyone had at LEAST 1. Here there's no such rule.
Show solution
Approach: Counterexample — pigeonhole forces a big count, not a tie
The claim is NOT true in general. More leaves than trees only forces some tree to have several leaves; it does not force two trees to have the EXACT SAME count.
Counterexample: 2 trees and 3 leaves (more leaves than trees). Give one tree 0 leaves and the other 3 leaves — more leaves than trees, yet the two counts differ.
Why did the 'friends' problems work but this doesn't? There, everyone had at least 1 friend, squeezing the counts into the \(N-1\) values \(1, \dots, N-1\) — one fewer box than people. Here 0 leaves is allowed, so the counts aren't squeezed and the trick fails.
Counting & ProbabilityLogic & Word Problemsand-process-multiplylogical-reasoning
Blake thinks his chance of getting into college A is 0.75 and into college B is 0.5. He multiplies to claim the chance of getting into BOTH is 0.375. Explain why his reasoning might not be right.
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Answer: Multiplying assumes independence, which is doubtful here
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Hint 1 of 3
Multiplying \(P(A) \times P(B)\) for 'A and B' only works under one special condition.
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Hint 2 of 3
What has to be true about the two events for multiplying to be valid?
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Hint 3 of 3
Think about whether getting into one college is really unrelated to getting into the other (his grades, scores, and essays affect both).
Multiplying two probabilities to get 'both happen' only works when the two events are INDEPENDENT — one happening has no effect on the chance of the other.
Blake's arithmetic (\(0.75 \times 0.5 = 0.375\)) is fine, but the multiplication itself is only allowed if the two acceptances are independent.
In real life they probably are NOT: the same grades, test scores, and essays affect both colleges. A strong applicant tends to get into both, a weaker one tends to be rejected by both.
So the events are linked, and the true chance of 'both' is likely higher than 0.375. His reasoning isn't justified unless the events are independent, which is doubtful here.
The integers 1 through 25 are arbitrarily separated into five groups of 5 numbers each. The median of each group is found, and M is the median of those five medians. What is the least possible value of M?
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Answer: A — 9.
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Hint 1 of 2
Unpack what M even is: it's the median of the five medians, i.e. the 3rd-smallest of them. To push M down, you need three groups whose medians are all small at once.
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Hint 2 of 2
Here's the bottleneck: a group's median needs two strictly smaller numbers sitting below it. Three small medians plus their six ‘below’ numbers eat up a lot of the tiny values — and tiny values are scarce. Count how many you'd need.
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Approach: lower-bound by counting scarce small numbers, then build a matching example
First decode M: it is the 3rd-smallest of the five group medians. So making M tiny requires three groups to each have a small median simultaneously.
Now the scarcity argument. Each of those three medians needs two numbers strictly below it inside its group. Suppose M ≤ 8. Then three medians (each ≤ 8) together with their six below-numbers (each smaller still) are 9 distinct values all ≤ 8 — impossible, since only 8 numbers (1–8) are that small. So M ≥ 9.
Then show 9 is actually reachable: {1, 2, 7, 24, 25}, {3, 4, 8, 22, 23}, {5, 6, 9, 20, 21} have medians 7, 8, 9, while the last two groups {10–14} and {15–19} absorb the big numbers (medians 12, 17). The five medians 7, 8, 9, 12, 17 have median 9.
Bound met by an example ⇒ the least value is 9.
Why this transfers: extremal problems are won by pairing a lower bound (a counting/pigeonhole reason it can't be smaller) with a construction (an explicit example hitting that bound). Neither half alone is a proof — you need both jaws of the vise.
The hardest pressure is where many pods are mutually connected — tackle that knot first. Find the largest group of pods that are all directly linked to each other; their grades are squeezed into a tiny set.
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Hint 2 of 2
Pods A, B, C, F are pairwise connected (a 4-clique), so their four grades must be mutually ≥ 2 apart. From {1,…,7}, the only such quadruple is {1, 3, 5, 7} — the most spread-out choice.
Show solution
Approach: find the clique to lock down four grades, then radiate outward
Spot the tightest constraint: A, B, C, F are all mutually connected, so their four grades must each differ by ≥ 2. Packing 4 grades into 1–7 that far apart forces exactly {1, 3, 5, 7} — there's no slack.
G connects to A and F. If G = 2, then A, F ≠ 1 or 3, forcing {A, F} ⊂ {5, 7} and {B, C} = {1, 3}.
D and E only touch C and F. The extreme grades 1 and 7 each have just one neighbor in this clique, so place 1 at C and 7 at F. The remaining {4, 6} go to D, E.
Filling in: D = 6 (avoids 7 enough), E = 4 (avoids 1 and 7), B = 3 (next to C = 1), A = 5. All constraints hold.
C + E + F = 1 + 4 + 7 = 12.
Why this transfers: in any constraint/coloring puzzle on a network, attack the densest cluster (the clique) first — it has the fewest possibilities, so it locks in the most and leaves the loose pods easy to finish.
Kaleana shows her test score to Quay, Marty, and Shana, but the others keep theirs hidden. Quay thinks, "At least two of us have the same score." Marty thinks, "I didn't get the lowest score." Shana thinks, "I didn't get the highest score." List the scores from lowest to highest for Marty (M), Quay (Q), and Shana (S).
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Answer: A — S, Q, M.
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Hint 1 of 2
The crucial detail is what each person can SEE. Quay only saw Kaleana's score, yet he's CERTAIN two scores match — how could he be certain knowing just one other score?
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Hint 2 of 2
He can only be sure if his own score equals the one he saw: Q = K. Once that's locked, turn Marty's and Shana's thoughts into comparisons with K.
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Approach: turn each statement into an inequality
Quay is certain two scores are equal but has seen only Kaleana's — the sole way to be sure is if HIS score matches it. So Q = K.
Marty knows he's not lowest, and the only score he's seen is Kaleana's, so M > K. Shana knows she's not highest, so S < K. Swap K for Q: S < Q < M.
Lowest to highest: S, Q, M. The transferable move in "who-knows-what" logic: a person's certainty is built only from what they can actually observe — that constraint is what pins everything down.
Every triangle on the top half lands on exactly one on the bottom — nothing vanishes. So just track one half and watch each color get used up.
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Hint 2 of 2
Work color by color: count how many reds and blues are "spent" by the given pairs, see what's forced to pair with white, and the leftover whites pair with each other.
Show solution
Approach: account for each color on one half (conservation)
One half has 3 red, 5 blue, 8 white. The 2 red-red pairs spend 2 reds, leaving 1 red; the 3 blue-blue pairs spend 3 blues, leaving 2 blue.
Now the leftovers must pair with whites. The 2 red-white pairs spend that last red and 1 white. The 2 leftover blues can't make a 4th blue-blue pair (only 3 are allowed), so each pairs with a white — using 2 more whites.
Whites spent: 1 (with red) + 2 (with blue) = 3, leaving 8 − 3 = 5 whites per half. Those face each other as 5 white-white pairs.
The engine here is conservation: a folded figure pairs everything one-to-one, so once you know how each "colored" piece is consumed, the remainder of any one color is forced.
You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $1.02, with at least one coin of each type. How many dimes must you have?
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Answer: A — 1 dime.
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Hint 1 of 2
'At least one of each' is a gift: spend one of each coin up front (41¢) and the puzzle shrinks to placing the 5 leftover coins. Then look at the LAST digit of what's left.
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Hint 2 of 2
The total ends in 2 and only pennies change the units digit. So the number of pennies is forced by the final digit — pin that down before worrying about the bigger coins.
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Approach: pay one of each first, then let the units digit fix the pennies
Take one of each coin off the top: 1 + 5 + 10 + 25 = 41¢, using 4 of the 9 coins. Remaining: 102 − 41 = 61¢ in 5 more coins.
Units digit move: nickels, dimes, quarters all end in 0 or 5, so only pennies can produce the final '1' in 61. With 5 coins, you can't afford 6 pennies — so exactly 1 more penny. Now 60¢ in 4 coins.
60¢ in 4 coins needs a quarter (four dimes max out at 40¢). Two quarters = 50¢ leaves 10¢ in 2 coins = two nickels. That fills all four with no dime.
So beyond the single starter dime, none are added: 1 dime.
The reusable trick: in coin/total problems, read the *last digit* of the amount — only pennies (the 1¢ pieces) can change it, so the units digit alone often pins the smallest-coin count and collapses the casework.
Logic & Word ProblemsNumber Theorywork-backwardaccount-for-all-possibilities
Same take-away game (pile of \(27\), take \(1\) to \(4\) each turn). New rule: when all counters are gone, the player who has collected an EVEN number of counters wins. (This is harder — you must keep track of both how many you leave AND whether your own pile is even or odd.) What should your first move be?
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Answer: Go first and take 2
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Hint 1 of 4
This time your position depends on two things at once: how many counters you leave, and whether your own collected pile is even or odd. Track both.
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Hint 2 of 4
Start at the end. If you already hold an EVEN number and you can leave \(0\) or \(1\), you win. So aim to finish holding an even amount.
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Hint 3 of 4
If you hold an ODD number and leave \(5\), check all four replies: they take \(1\), you take \(3\); take \(2\), you take \(3\); take \(3\), you take \(1\); take \(4\), you take \(1\). In every case your pile flips to EVEN and you leave a deadly \(0\) or \(1\).
Show solution
Approach: Working backward while tracking two states (counters left and parity of your pile)
Here you win based on whether YOUR OWN pile is even at the end, so you track two things: the number you leave, and the even/odd state of what you hold.
When you hold an ODD number, leaving \(5\) wins. Check every reply: take \(1\) then you take \(3\); take \(2\) then you take \(3\); take \(3\) then you take \(1\); take \(4\) then you take \(1\). Each time you add enough so your pile becomes EVEN and you leave a fatal \(0\) or \(1\).
When you hold an EVEN number, the safe leaves come in pairs. The base pair is \(0, 1\) (you already won) and the next pair is \(6, 7\). Continuing by working backward gives — even pile: leave \(0, 1, 6, 7, 12, 13, 18, 19, 24, 25\); odd pile: leave \(5, 11, 17, 23\).
From \(27\), the one good first move is to take \(2\): now you hold \(2\) (even) and have left \(25\), which is on the even list. After that, always land on the list matching your current parity.
Number TheoryLogic & Word Problemssolve-a-simpler-problempattern-recognition
People stand in a circle, numbered \(1, 2, 3, \dots, N\) clockwise. Person \(1\) says 'one' and stays. Person \(2\) says 'two' and steps OUT. Person \(3\) stays, person \(4\) steps out, and so on around and around: every other person leaves. The counting keeps going (it does NOT restart) until one person is left. Who survives when \(N = 100\)?
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Answer: Person 73
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Hint 1 of 4
One hundred is too many to act out. Solve smaller versions first! Make a table: for \(N = 1, 2, 3, 4, 5, 6, 7, 8\), who survives?
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Hint 2 of 4
Notice the rows where the survivor is person \(1\). They happen at \(N = 1, 2, 4, 8, 16, \dots\) — the powers of \(2\). What pattern do you see between those points?
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Hint 3 of 4
Write \(N\) as (a power of \(2\)) plus a leftover: \(N = 2^k + L\), where \(2^k\) is the biggest power of \(2\) that is not bigger than \(N\). The survivor turns out to be a simple formula in \(L\).
Show solution
Approach: Solve a simpler problem, find the pattern, then apply the formula
Build a table of survivors for small \(N\):
N
1
2
3
4
5
6
7
8
survivor
1
1
3
1
3
5
7
1
The survivor is person \(1\) exactly when \(N\) is a power of \(2\) (\(1, 2, 4, 8, 16, \dots\)). Between powers of \(2\), the survivor jumps up by \(2\) each time (\(1, 3, 5, 7, \dots\)) and then resets to \(1\).
So write \(N = 2^k + L\), where \(2^k\) is the biggest power of \(2\) that fits inside \(N\) and \(L\) is the leftover. The survivor's number is \(2L + 1\).
For \(N = 100\): the biggest power of \(2\) that is \(\le 100\) is \(64\), so \(L = 100 - 64 = 36\) and the survivor is \(2(36) + 1 = 73\).
Logic & Word ProblemsCounting & Probabilityaccount-for-all-possibilitiesorganizing-datareduce-and-expandchoosing-vs-arranging
A captured hero must design a new flag with three vertical stripes, each stripe a different color. He can only use colors that already appear on these five real three-stripe flags, which together use \(6\) different colors: Belgium (black, yellow, red); France (blue, white, red); Ireland (green, white, yellow); Italy (green, white, red); Romania (blue, yellow, red). How many different three-stripe flags can be made from these \(6\) colors? How many of those are brand new (not one of the five that already exist)?
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Answer: 120 possible flags in all; 115 of them are new
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Hint 1 of 4
There are really two questions hiding here: WHICH three colors to use, and in WHAT order to put them left to right. Solve them one at a time.
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Hint 2 of 4
First count how many ways to choose \(3\) different colors out of \(6\) when order doesn't matter yet. List them carefully so you don't miss any.
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Hint 3 of 4
There are \(20\) ways to choose the three colors. Now take one set of three colors and count the left-to-right orders: ABC, ACB, BAC, BCA, CAB, CBA. How many is that?
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Approach: Separate choosing the colors from arranging them, then subtract existing flags
Step 1 — choose the three colors: pick 3 of the 6 with order not mattering yet; there are exactly 20 sets (\(\binom{6}{3}=\dfrac{6\times5\times4}{3\times2\times1}=20\)).
Step 2 — order each set: three chosen colors A, B, C arrange left-to-right as ABC, ACB, BAC, BCA, CAB, CBA — exactly 6 (\(3!=6\)). Since stripes are ordered, each order is a different flag.
Step 3 — total flags: \(20\times6=120\). (Directly: 6 choices for the first stripe, 5 for the second, 4 for the third, \(6\times5\times4=120\).)
Step 4 — new flags: remove the 5 that already exist: \(120-5=115\) brand-new flags.
Ratios, Rates & ProportionsLogic & Word Problemslogical-reasoningvisual-representationpattern-recognition
Anton and Ben start running toward each other from the two ends of a long straight path (Anton from end A, Ben from end B). Each runs at his own steady speed. They first meet 800 m from Ben's end. They keep going, reach the far ends, instantly turn around, and meet a second time 400 m from Anton's end. How long is the path (in meters)?
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Answer: 2000 m
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Hint 1 of 4
Don't track each runner separately at first — track the TOTAL distance the two of them run together.
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Hint 2 of 4
At the FIRST meeting, the two of them together have covered the path exactly once (their two pieces fill it).
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Hint 3 of 4
At the SECOND meeting, together they have covered the path exactly three times. (Draw it: each one went to the far end and partway back.)
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Approach: Combined distance triples between the two meetings
Because both run steadily, their distances grow in the same proportion. At the first meeting they have together run one full path length; at the second meeting they have together run three path lengths (each finishes the path and comes partway back).
So the combined distance tripled, and since both run steadily, each runner's own distance also triples.
Ben ran 800 m to the first meeting (it was 800 m from his end B), so by the second meeting Ben has run \(3 \times 800 = 2400\) m. The second meeting is 400 m from Anton's end A, meaning Ben ran the whole path and came back 400 m, so path \(= 2400 - 400 = 2000\) m.
The path is 2000 m long. (Check Anton: he ran \(2000 - 800 = 1200\) m to the first meeting, then \(3 \times 1200 = 3600\) m by the second, which is one path plus 1600 m back, leaving him \(2000 - 1600 = 400\) m from end A. It fits.)
Logic & Word ProblemsCounting & ProbabilityNumber Theoryaccount-for-all-possibilitiesorganizing-datapattern-recognitionvisual-representation
A knight starts on the bottom-left square of a standard \(8 \times 8\) chessboard. Each knight move always goes upward, climbing either \(1\) row or \(2\) rows (never going down). Call a move that climbs \(2\) rows 'long' and a move that climbs \(1\) row 'short.' (We don't care about left-or-right, only the rows.) (a) How many rows must the knight climb to reach the top row? (b) List every combination of long and short moves that does it. (c) Stretch: for each combination, how many different orders are there to make those moves? Add them up.
The board has \(8\) rows. If the knight starts on row \(1\) (the bottom) and must reach row \(8\) (the top), how many rows does it gain in total?
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Hint 2 of 4
Let \(a\) = number of long moves (2 rows each) and \(b\) = number of short moves (1 row each). The climb adds up to \(7\) rows. Write an equation connecting \(a\) and \(b\).
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Hint 3 of 4
From \(2a + b = 7\): since \(2a\) is even and \(7\) is odd, \(b\) must be odd. List the pairs \((a,b)\) where \(a\) and \(b\) are \(0\) or more.
Show solution
Approach: Set up 2a+b=7, list whole-number solutions, then count arrangements
(a) Going from row 1 to row 8 is a gain of \(8-1=7\) rows.
(b) Let \(a\) be long moves (2 rows) and \(b\) short moves (1 row), so \(2a+b=7\). Since \(2a\) is even and 7 is odd, \(b\) is odd. The solutions are \((a,b)=(3,1),(2,3),(1,5),(0,7)\) — 4 combinations.
(c) Count arrangements for each: \((3,1)\) has 4 moves, the single short in any of 4 spots → 4; \((2,3)\) has 5 moves, choose 2 long → \(\dfrac{5\times4}{2\times1}=10\); \((1,5)\) has 6 moves, single long in any of 6 spots → 6; \((0,7)\) → 1.
Adding: \(4+10+6+1=21\) different upward move-patterns.
Logic & Word ProblemsGeometry & Measurementconsidering-extreme-caseslogical-reasoningvisual-representation
It is raining straight down, steadily, with no wind. You need to get from point \(C\) to point \(P\). To stay as dry as possible, should you run, walk slowly, or does it not matter?
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Answer: Run — the front catches the same rain at any speed, but the top of your head catches less the faster you go
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Hint 1 of 4
Split your wetness into two parts: rain landing on the TOP of your head, and rain hitting your FRONT as you move forward. Think about each part separately.
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Hint 2 of 4
Top of your head: the longer you are out in the rain, the more drops land on top. So going faster (less time outside) means less rain on top.
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Hint 3 of 4
Your front: as you move forward you 'sweep up' all the drops in your path. Whether you go fast or slow, you still pass through the same stretch of air from \(C\) to \(P\), so your front meets the same number of drops either way.
Show solution
Approach: Considering extreme cases — split the wetness into top and front
Break the problem into two pieces and think about extremes.
Top of the head: imagine walking infinitely slowly — you stand in the rain forever and your head gets soaked. Imagine running infinitely fast — you are barely outside, so almost nothing lands on top. So faster means less rain on top.
Front of the body: as you travel from \(C\) to \(P\), you run into all the raindrops in the space you pass through. That space is the same no matter how fast you cross it, so your front collects the same amount of rain at any speed.
Putting them together: the front is a tie, and the top favors speed. So running gets you to \(P\) driest overall.
Logic & Word ProblemsCounting & ProbabilityArithmetic & Operationslogical-reasoningorganizing-datapattern-recognition
A senator must meet five groups, one at a time; while a group is being seen, everyone in later groups waits. Group 1 has 4 members and meets 20 min; group 2 has 8 members, 10 min; group 3 has 5 members, 30 min; group 4 has 10 members, 15 min; group 5 has 6 members, 25 min. In what order should he call the groups to make the total waiting time of all people as small as possible?
This is like the dentist problem, but each 'patient' is a whole group of people who all wait together.
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Hint 2 of 4
A group that takes a long time but has few people isn't so bad. What matters is the time PER PERSON in the group.
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Hint 3 of 4
Compute time divided by members for each group.
Show solution
Approach: Schedule by smallest time-per-member first
A group of \(g\) people meeting for \(t\) minutes acts like \(g\) people who each weigh \(t/g\) minutes. So compare time per member, then go smallest-first.
So the best order is G2, G4, G5, G1, G3. (Swapping any neighbor pair shows the earlier group should have the smaller time-per-person, which forces this whole order.)
Counting & ProbabilityLogic & Word Problemspigeonholelogical-reasoningsymmetry
A round table has 5 chairs, and a name card is taped to the table in front of each chair. Five friends sit down without looking, and it turns out nobody is in front of their own card. Show that you can spin the table to a new position where at least 2 friends end up in front of their own cards.
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Answer: some spin makes at least 2 friends correct
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Hint 1 of 4
Spinning the table one chair at a time moves every card forward one seat. There are 4 'new' spin positions (after 1, 2, 3, or 4 clicks); the starting position has nobody correct.
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Hint 2 of 4
Pick one friend. As the table makes its 4 clicks, that friend's own card passes by their seat exactly once. So exactly one spin position is 'correct' for that friend.
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Hint 3 of 4
Every friend has exactly one correct spin position among the 4. Make the 4 spin positions your boxes, and assign each friend to their correct one.
Spin the table one click at a time. There are 4 'new' positions (1, 2, 3, or 4 clicks); the no-spin start has nobody correct, by assumption.
Focus on one friend. As the cards parade past that friend's seat over the 4 clicks, the friend's own card lines up with the seat exactly once (not at the start, so during one of the 4 clicks). So each friend has exactly one correct spin position.
Make the 4 spin positions our boxes and put each of the 5 friends into the box for their one correct spin. Since \(5 > 4\), two friends land in the same box.
At that single spin position, both of those friends sit in front of their own cards — so some spin makes at least \(2\) friends correct.
Counting & ProbabilityLogic & Word Problemspigeonholecasework
Color every square of a \(3 \times 9\) grid either black or white, any way you like. Show that you can always find a rectangle (using 2 of the rows and 2 of the columns) whose 4 corner squares are all the same color.
Show answer
Answer: a same-color-corner rectangle always exists
Show hints
Hint 1 of 4
A rectangle's 4 corners live in 2 columns and 2 rows. Try looking at the grid one column at a time — what can you always say about a single column of 3 squares in 2 colors?
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Hint 2 of 4
How many 'features' are possible? Choose 2 of the 3 rows (there are 3 ways: rows 1&2, 1&3, 2&3) and a color (2 ways). That's \(3 \times 2 = 6\) features — your boxes.
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Hint 3 of 4
You have 9 columns, each giving a feature, but only 6 feature-boxes.
Show solution
Approach: Pigeonhole on (row-pair, color) features — 9 columns, 6 features
Look at any single column: 3 squares in 2 colors, so by pigeonhole at least 2 of its squares share a color. For each column record a 'feature': which two rows match, and what color they share.
How many features are possible? 3 ways to pick the matched row-pair (1&2, 1&3, or 2&3) times 2 color choices, so \(3 \times 2 = 6\) features. Make these 6 features the boxes.
The grid has 9 columns, each giving one feature, but only 6 boxes. Since \(9 > 6\), two columns share a feature.
That means the same two rows have the same color in both columns — those 4 squares are the corners of a rectangle, all one color.
Logic & Word ProblemsCounting & Probabilityvisual-representationlogical-reasoning
A building project has stages \(P_1, \dots, P_6\) (\(P_1\) start, \(P_6\) finish). An arrow '\(P_2 \to P_5 = 9\)' means \(P_2\) must finish before \(P_5\) and that step takes 9 days. The steps (in days): \(P_1\to P_2 = 4\), \(P_2\to P_3 = 6\), \(P_2\to P_5 = 9\), \(P_2\to P_4 = 8\), \(P_3\to P_5 = 7\), \(P_5\to P_6 = 3\), \(P_4\to P_6 = 6\). What is the shortest total time to finish the whole project (in days)?
Show answer
Answer: 20 days
Show hints
Hint 1 of 4
Draw the stages as dots and each step as an arrow with its days. Steps that don't depend on each other can happen at the same time.
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Hint 2 of 4
Surprise: the project isn't done until EVERY chain of steps is done. So the finish time is set by the LONGEST chain from start to finish.
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Hint 3 of 4
List every path of arrows from \(P_1\) to \(P_6\) and add up the days along each one.
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Approach: Longest path (critical path) in the activity network
Steps on different branches run at the same time, so the project finishes only when the longest must-do-in-order chain is complete.
List the paths from \(P_1\) to \(P_6\): \(P_1\to P_2\to P_3\to P_5\to P_6 = 4+6+7+3 = 20\); \(P_1\to P_2\to P_5\to P_6 = 4+9+3 = 16\); \(P_1\to P_2\to P_4\to P_6 = 4+8+6 = 18\).
The longest is 20 days, along \(P_1\to P_2\to P_3\to P_5\to P_6\), which is the bottleneck (critical path).
So the whole project needs 20 days; the other branches finish within that time.
Logic & Word ProblemsGeometry & Measurementproof-by-contradictionpigeonhole
Six round disks lie in the plane. No disk contains the center of any other disk. Prove that the six disks cannot all share a single common point.
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Answer: Proven impossible: a shared point forces two centers within 60°, so one disk would contain another's center
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Hint 1 of 4
Argue by contradiction: pretend all six disks DO share a common point \(P\). Draw a segment from \(P\) out to each of the six centers.
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Hint 2 of 4
The six segments spread out around \(P\), and the angles around a point add up to \(360^\circ\). With six angles sharing \(360^\circ\), at least one angle is \(60^\circ\) or smaller (pigeonhole: they can't all be bigger than \(60\)).
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Hint 3 of 4
Take the two centers making that small angle. Call their distances from \(P\) \(a\) and \(b\), with \(a\) the smaller. In that skinny triangle, the side across from the small (\(\le 60^\circ\)) angle is not the longest side.
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Approach: Proof by contradiction with pigeonhole on the angles around a point
Suppose, for contradiction, that all six disks share a common point \(P\). Draw the six segments from \(P\) to the centers \(O_1,\dots,O_6\).
The angles around \(P\) total \(360^\circ\). If all six were bigger than \(60^\circ\) they'd exceed \(360^\circ\), so by pigeonhole at least one angle is \(\le 60^\circ\). Call its two centers \(O_a\) and \(O_b\), at distances \(a\le b\) from \(P\).
In triangle \(PO_aO_b\) the angle at \(P\) is \(\le 60^\circ\), so it isn't the largest angle, and the side opposite it, \(O_aO_b\), isn't the longest side; hence \(O_aO_b\le b\).
Since \(P\) is inside disk \(b\), \(b\) is at most that disk's radius, so \(O_aO_b\le b\le(\text{radius of disk }b)\). That puts center \(O_a\) inside disk \(b\) — one disk contains another's center, contradicting the rule. So the six disks cannot all share a common point.
Counting & ProbabilityLogic & Word Problemsand-process-multiplylogical-reasoning
A state uses plates of 3 letters followed by 3 digits, and every possible plate is already used. To make more plates cheaply, it will add ONE more character slot. Should the new slot be a letter or a digit to create the most new plates? Explain.
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Answer: Add a letter (26 choices)
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Hint 1 of 3
Adding one slot multiplies the number of plates by however many choices that slot has. So you want the slot with the most choices.
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Hint 2 of 3
A letter slot has 26 choices; a digit slot has only 10. Which multiplies your plates more?
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Hint 3 of 3
Compare multiplying by 26 versus multiplying by 10.
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Approach: Adding a slot multiplies by the slot's choice count
Adding one extra slot multiplies the number of plates by the number of symbols that slot allows. So to make the MOST new plates, choose the slot with the most options.
A letter offers 26 choices; a digit offers only 10. Since 26 is bigger than 10, adding a LETTER slot makes far more plates.
The old pool was \(26^3 \times 10^3 = 17{,}576{,}000\) plates. A letter multiplies by 26, giving \(456{,}976{,}000\); a digit would only multiply by 10, giving \(175{,}760{,}000\). So add a letter.
Logic & Word Problemslogical-reasoningaccount-for-all-possibilities
Four glasses sit at the corners of a square table, each right-side-up or upside-down. You are blindfolded. On each turn you may feel any two glasses and flip none, one, or both. After each turn the table is spun randomly, so you lose track of corners. A bell rings the instant all four glasses face the same way. What is the fewest number of turns that GUARANTEES you can make the bell ring?
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Answer: 5 turns
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Hint 1 of 4
You can't control which corners you grab compared to last time (the spin scrambles them), but you CAN choose two ADJACENT (side-by-side) glasses or two DIAGONAL (across) glasses. Mix the two kinds.
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Hint 2 of 4
Turn 1: grab a diagonal pair and turn both right-side-up. Turn 2: grab an adjacent pair and turn both right-side-up. If no bell yet, exactly one glass is upside-down.
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Hint 3 of 4
Turn 3: grab a diagonal pair and flip BOTH. If no bell, the two down glasses are now adjacent (side by side).
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Approach: Mix diagonal and adjacent grabs to force all four to match
Mix DIAGONAL and ADJACENT grabs so that no matter how the table spins, the glasses are forced toward all-the-same.
Turn 1 (diagonal): set both right-side-up. Turn 2 (adjacent): set both right-side-up. If no bell, exactly one glass is down.
Turn 3 (diagonal): flip both. If no bell, the two down glasses are now adjacent. Turn 4 (adjacent): flip both. If no bell, the two down glasses are now diagonal.
Turn 5 (diagonal): flip both — now all four match and the bell rings. Every branch finishes by turn \(5\), and no faster plan is guaranteed, so the answer is \(5\) turns.
Five runners, P, Q, R, S, T, have a race. P beats Q, P beats R, Q beats S, and T finishes after P and before Q. Who could NOT have finished third in the race?
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Answer: C — P and S.
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Hint 1 of 2
Don't try to build one full finishing order — there are many. Instead, link the clues into a chain (who-beats-whom) and ask of each runner: how many people MUST be ahead of them, and how many MUST be behind?
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Hint 2 of 2
To be exactly third, a runner needs room: at least 2 people forced ahead AND at least 2 forced behind. Anyone with too many forced ahead (or too many forced behind) is locked out of 3rd.
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Approach: count forced-ahead and forced-behind for each runner
Chain the clues. 'T after P and before Q' plus 'Q beats S' gives P < T < Q < S, and separately P < R. Read positions as left-to-right finishing order.
P sits ahead of T, Q, R, and S — all four others — so P is forced into 1st. With 0 people who can be ahead of it, P can never be 3rd.
S sits behind Q, which is behind both P and T, so at least P, T, Q finish ahead of S: that's 3 runners, pinning S at 4th or 5th. S can never be 3rd either.
Everyone else (Q, R, T) has enough wiggle room to land 3rd in some valid order, so the runners who could NOT be third are P and S.
Why this transfers: for ranking-with-clues, you rarely need the full order. Turn the clues into one chain, then count must-be-ahead / must-be-behind. A position is impossible for anyone whose forced-ahead count is too big (can't move up) or too small (forced even higher).
Alan, Beth, Carlos, and Diana were discussing their possible grades in mathematics class this grading period. Alan said, "If I get an A, then Beth will get an A." Beth said, "If I get an A, then Carlos will get an A." Carlos said, "If I get an A, then Diana will get an A." All of these statements were true, but only two of the students received an A. Which two received A's?
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Answer: C — Carlos, Diana.
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Hint 1 of 3
Picture the three statements as falling dominoes pointing one way: Alan → Beth → Carlos → Diana. If a domino early in the line falls (gets an A), every domino after it must fall too. So who can afford to get an A without toppling too many?
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Hint 2 of 3
Count the dominoes that fall from each starting point. Alan's A topples 4 in all; Beth's topples 3; only starting near the *end* of the chain keeps the total at exactly 2.
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Hint 3 of 3
The arrows only point forward, so getting an A never forces anyone *earlier* — that's why the two A-getters must be the last links.
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Approach: follow the forward-only domino chain
Each 'if I get an A, then the next person does' is a one-way arrow: Alan → Beth → Carlos → Diana. An A anywhere knocks down everyone *after* it, but never anyone before.
Test from the front. If Alan gets an A, the chain forces Beth, Carlos, and Diana too — that's 4 A's, too many. If Beth gets an A, it forces Carlos and Diana — 3 A's, still too many. So neither Alan nor Beth can be one of the two.
That leaves the tail end. If Carlos gets an A, only Diana is forced — exactly 2 A's, and nobody else is dragged in: Carlos and Diana.
Why start from the front: because the arrows point forward, the *earlier* a person is, the more A's they trigger. To keep the count smallest, the A's must sit as far down the chain as possible — a handy rule for any 'if-then' chain problem.
Logic & Word Problemscontrapositivefind-counterexample
Five cards are lying on a table as shown.
PQ
346
Each card has a letter on one side and a whole number on the other side. Jane said, “If a vowel is on one side of any card, then an even number is on the other side.” Mary showed Jane was wrong by turning over one card. Which card did Mary turn over?
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Answer: A — 3.
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Hint 1 of 3
The rule is one-directional: 'vowel → even'. The ONLY way to break it is to find a single card that is a vowel on one side and an ODD number on the other. So ask: which visible faces could possibly be hiding that forbidden pairing?
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Hint 2 of 3
Sort the five visible faces. P and Q are consonants — the rule says nothing about consonants, so flipping them proves nothing. 4 and 6 are even — even is exactly what the rule allows, so they're safe too. That leaves only the odd-numbered face.
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Hint 3 of 3
To catch a violation, flip the card whose visible side is ODD: if a vowel is hiding behind it, the rule is busted. Which card is that?
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Approach: test only the cards that could falsify the implication
Breaking 'vowel → even' requires a card that's a vowel paired with an odd number. Check what flipping each face could reveal: P, Q (consonants) — the rule doesn't restrict consonants, useless to flip; 4, 6 (even) — the rule permits any letter behind an even number, also useless; 3 (odd) — if a vowel hides behind it, that's a vowel with an odd number, a direct contradiction.
Only the odd-numbered card can expose the forbidden vowel-with-odd combination, so Mary turned over 3.
Why this transfers: to test an 'if A then B' rule, you only ever check the A-cases (is B true?) and the not-B-cases (is A false behind it?). Cases that are already not-A or already B can never disprove it — a sharp filter that saves you from flipping every card.
