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On a real national test, more than half of junior-high students missed this question: What is 75% of 12? That is surprising, because the numbers are so friendly! Show how to see the answer with a picture instead of just punching buttons. (Then try the deliberately unfriendly version: what is 74% of 13?)
Show answer
Answer: 9 (and 74% of 13 is about 9.6)
Show hints
Hint 1 of 4
The word 'percent' just means 'out of 100.' Is there a simple fraction that equals 75%?
Still stuck? Show hint 2 →
Hint 2 of 4
Draw 12 little squares. If you split them into 4 equal groups, how many squares are in each group?
Still stuck? Show hint 3 →
Hint 3 of 4
75% is the same as 3 out of every 4, which is the fraction \(\tfrac34\). Take \(\tfrac14\) of 12 first, then take 3 of those groups.
Show solution
Approach: See the percent as a friendly fraction and picture it
The trick is to see 75% as the friendly fraction \(\tfrac34\), not to reach for a percent rule.
Draw 12 squares in a 3-by-4 array and split them into 4 equal columns. Each column has 3 squares, so each column is \(\tfrac14\) of the whole.
\(\tfrac14\) of 12 = 3 squares (one column), so \(\tfrac34\) of 12 = three columns = 3 + 3 + 3 = 9.
So 75% of 12 = \(\tfrac34 \times 12 = 9\).
The unfriendly twin 74% of 13 looks almost the same on paper, but 74% is not a clean fraction and 13 won't split into equal small groups, so there is no neat picture — you would just estimate \(0.74 \times 13 \approx 9.6\). The real lesson: grab the easy picture when the numbers are friendly.
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)
Show hints
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.
Still stuck? Show hint 2 →
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?
Still stuck? Show hint 3 →
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 \(7\)? (\(11-7\).)
Still stuck? Show hint 2 →
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!
A group of children riding on bicycles and tricycles rode past Billy Bob's house. Billy Bob counted 7 children and 19 wheels. How many tricycles were there?
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Answer: C — 5 tricycles.
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Hint 1 of 2
Don't set up two equations — pretend every child is on a bicycle first and watch what's missing.
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Hint 2 of 2
Each tricycle is just a bicycle with one extra wheel, so the leftover wheels count the tricycles directly.
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Approach: assume all bicycles, then the leftover wheels count the tricycles
Suppose all 7 children rode bicycles. That would be 7 × 2 = 14 wheels — but only counts as a starting guess.
We actually see 19 wheels, so 19 − 14 = 5 wheels are unaccounted for. A tricycle is just a bicycle with one extra wheel, so each leftover wheel marks one tricycle: 5 tricycles.
You'll see this again: the "assume the cheapest option, then spend the surplus" trick cracks chickens-and-rabbits, coins, and stamp problems without any algebra.
Another way — solve the system:
With b bicycles and t tricycles: b + t = 7 and 2b + 3t = 19.
Subtract twice the first from the second: t = 19 − 14 = 5.
A hallway has \(100\) lockers, numbered \(1\) to \(100\), all closed. Student \(1\) opens every locker. Student \(2\) changes (opens if closed, closes if open) every \(2\)nd locker: \(2, 4, 6, \dots\). Student \(3\) changes every \(3\)rd locker: \(3, 6, 9, \dots\). This continues through student \(100\). After all \(100\) students go by, how many lockers are OPEN?
Show answer
Answer: 10 lockers (the perfect squares)
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Hint 1 of 4
Pick one locker, say number \(12\), and figure out exactly which students touch it. A student numbered \(n\) touches locker \(12\) only if \(n\) divides \(12\).
Still stuck? Show hint 2 →
Hint 2 of 4
So locker \(m\) gets touched once for each divisor of \(m\). A locker ends OPEN if it was touched an ODD number of times. So the question becomes: which numbers have an odd number of divisors?
Still stuck? Show hint 3 →
Hint 3 of 4
Divisors usually come in pairs. For \(12\): \((1, 12), (2, 6), (3, 4)\) — that is \(6\) divisors, an even number, so locker \(12\) ends closed. When does a divisor get paired with ITSELF?
Show solution
Approach: Count divisors; perfect squares have an odd number of divisors
Focus on one locker, number \(m\). Student \(n\) touches it exactly when \(n\) divides \(m\). So locker \(m\) gets touched once for every divisor of \(m\), and it ends OPEN when the number of divisors is ODD.
Divisors come in pairs \((d, m/d)\). For example \(12\) has the pairs \((1,12), (2,6), (3,4)\) — six divisors, an even number, so locker \(12\) ends closed.
The only way to get an ODD count is when one divisor pairs with itself, meaning \(d = m/d\), so \(m = d^2\). For example \(9\) has divisors \(1, 3, 9\): the pair \((3,3)\) is just one number, giving the odd count \(3\).
So the lockers that stay open are exactly the perfect squares \(1, 4, 9, 16, 25, 36, 49, 64, 81, 100\) — ten lockers.
Many students freeze on \(5x=3x\) because there seems to be 'nothing' on the right. Solve \(5x=3x\) using the same first step you would use for \(5x=3x+6\).
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Answer: x = 0
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Hint 1 of 4
Do the very same first step you would do for \(5x=3x+6\): get all the \(x\)'s onto one side. Don't be thrown off by the missing number.
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Hint 2 of 4
Subtract \(3x\) from both sides. What is left on the left? What is left on the right?
Still stuck? Show hint 3 →
Hint 3 of 4
You get \(2x=0\). The right side is the NUMBER zero — a real number you can work with, not 'nothing.'
Show solution
Approach: Treat zero as a number and solve normally
Use the same move as for \(5x=3x+6\): subtract \(3x\) from both sides.
\(5x-3x=3x-3x\) gives \(2x=0\). The right side is the NUMBER zero, not 'nothing.'
Divide both sides by \(2\): \(x=0\).
So \(x=0\). The equation was never harder than the first one — it only felt that way because \(0\) can look like 'nothing.'
Mika wants to estimate how far a new electric bike goes on a full charge. She made two trips totaling 40 miles: the first used 12 of the battery and the second used 310 of the battery. How many miles can the bike go on a fully charged battery?
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Answer: C — 50 miles.
Show hints
Hint 1 of 2
The 40 miles didn't drain a full battery. First combine the two trips: what single fraction of the battery did the 40 miles actually use?
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Hint 2 of 2
Once you know 40 miles used some fraction of the charge, the rest is one proportion: scale that fraction up to a whole battery (the full charge is 5/5).
Show solution
Approach: find the fraction the 40 miles used, then scale to a whole battery
Combine the two trips: ½ + 3/10. With a common denominator, 5/10 + 3/10 = 8/10 = 4/5 of the battery powered the 40 miles.
If 4/5 of a charge gives 40 miles, each fifth gives 40 ÷ 4 = 10 miles, so a full 5/5 gives 5 × 10 = 50 miles.
Why this works: ‘a fraction of the whole equals a known amount’ is a proportion — find the value of one unit piece (here, one-fifth = 10 mi), then multiply up to the whole.
Another way — divide by the fraction:
4/5 of the battery = 40 miles, so the full battery is 40 ÷ 45 = 40 × 54 = 50 miles.
Never multiply all this out. Each fraction is (a number)/(that number + 2), so the lists of numerators and denominators are nearly the same — line them up and watch what cancels.
Still stuck? Show hint 2 →
Hint 2 of 2
Numerators: 1, 2, 3, …, 20. Denominators: 3, 4, …, 22. Everything from 3 to 20 appears in both and cancels — only the two ends survive.
Show solution
Approach: telescoping — the lists overlap, so almost everything cancels
Insight: don't compute — collect. Each factor is (k)/(k+2), so the numerators run 1, 2, 3, …, 20 and the denominators run 3, 4, 5, …, 22. Every number from 3 to 20 shows up in both lists, so it cancels.
Only the unmatched ends remain — 1 and 2 on top, 21 and 22 on the bottom: 1 · 221 · 22 = 2462 = 1231.
You'll see this again: a long product (or sum) where each term overlaps the next is a telescope — line up the lists and only the few leftover ends matter. The shift here is 2, so two terms survive on each end.
To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid (shown below). For the large kite she triples both the height and width of the entire grid.
The large kite is covered with gold foil. The foil is cut from a rectangular piece that just covers the entire grid. How many square inches of waste material are cut off from the four corners?
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Answer: D — 189 square inches.
Show hints
Hint 1 of 2
The kite's diagonals are exactly the rectangle's width and height — so kite area is ½ × width × height, which is half the rectangle. No need to measure the four corners separately.
Still stuck? Show hint 2 →
Hint 2 of 2
Waste = rectangle − kite, but since the kite IS half the rectangle, the waste is just the other half.
Show solution
Approach: waste = rectangle − kite, and the kite is half the rectangle
The large rectangle covers the whole grid: 18 × 21 = 378 square inches.
Here's the shortcut: the kite's diagonals span the full 18 and 21, so its area is ½ × 18 × 21 — exactly half the rectangle. The cut-off corners are therefore the other half: 378 ÷ 2 = 189 square inches.
Spotting that a shape is a clean fraction of its bounding box saves the messier work of computing four corner triangles one by one — look for that whenever a figure sits snugly inside a rectangle.
Another way — subtract the kite explicitly:
Kite area = ½ × 18 × 21 = 189 square inches.
Waste = rectangle − kite = 378 − 189 = 189 square inches — confirming the two halves match.
Ara and Shea were once the same height. Since then Shea has grown 20% while Ara has grown half as many inches as Shea. Shea is now 60 inches tall. How tall, in inches, is Ara now?
Show answer
Answer: E — 55 inches.
Show hints
Hint 1 of 2
The 60 inches is Shea's height AFTER a 20% growth — so 60 is 120% of the shared start, not the start itself. Undo the growth to find where they both began.
Still stuck? Show hint 2 →
Hint 2 of 2
Watch the units switch: Ara grew 'half as many *inches*,' not half the percent. Once you know Shea's inches gained, halve THAT number.
Show solution
Approach: undo the percent to find the start, then count inches
60 is 120% of the common starting height, so start = 60 ÷ 1.2 = 50 inches. That means Shea gained 60 − 50 = 10 inches.
Ara grew half as many *inches*: 10 ÷ 2 = 5 inches. Ara is now 50 + 5 = 55 inches.
The habit to build: the two people start equal but the problem mixes a percent (Shea) with raw inches (Ara). Always convert the percent into actual inches before comparing — reasoning in percents would have you compare 20% of one height with an inch count, which aren't the same kind of thing.
The Amaco Middle School bookstore sells pencils costing a whole number of cents. Some seventh graders each bought a pencil, paying a total of $1.43. Some of the 30 sixth graders each bought a pencil, and they paid a total of $1.95. How many more sixth graders than seventh graders bought a pencil?
Show answer
Answer: D — 4 more.
Show hints
Hint 1 of 2
The number of buyers in each grade must be a WHOLE number = (total paid) ÷ (one price). So the single pencil price has to divide BOTH totals evenly. Work in cents: 143 and 195.
Still stuck? Show hint 2 →
Hint 2 of 2
A number dividing both 143 and 195 is a common divisor — factor each and look for what they share.
Show solution
Approach: the price is a common divisor of both totals
Switch to cents so everything's whole: $1.43 = 143¢, $1.95 = 195¢. The price p must divide both (buyers = total ÷ p has to come out whole).
Factor: 143 = 11 × 13 and 195 = 3 × 5 × 13. Their only shared factors are 1 and 13.
Rule out p = 1¢: at 1¢ each, 195 sixth graders would have bought — but there are only 30. So p = 13¢.
Why this transfers: "equal items, total cost, unknown unit price" means the price divides every total — reach for common divisors (gcd), then use side conditions (like "at most 30 kids") to pick the right one.
Rohan keeps a total of 90 guppies in 4 fish tanks.
There is 1 more guppy in the 2nd tank than in the 1st tank.
There are 2 more guppies in the 3rd tank than in the 2nd tank.
There are 3 more guppies in the 4th tank than in the 3rd tank.
How many guppies are in the 4th tank?
Show answer
Answer: E — 26 guppies.
Show hints
Hint 1 of 2
The clues chain off each other, so anchor everything to ONE tank. From tank 1, each later tank is tank 1 plus a running total: +1, then +1+2 = +3, then +1+2+3 = +6.
Still stuck? Show hint 2 →
Hint 2 of 2
Technique: write all four as tank 1 + (0, 1, 3, 6). Their sum is 4·(tank 1) + 10 = 90 — one equation for one unknown.
Show solution
Approach: express every tank in terms of tank 1
The differences chain, so pin everything to tank 1 = x. Then tank 2 = x + 1, tank 3 = (x+1) + 2 = x + 3, tank 4 = (x+3) + 3 = x + 6.
Adding: 4x + (1 + 3 + 6) = 4x + 10 = 90, so x = 20.
The question wants tank 4, not tank 1 — so finish the job: tank 4 = 20 + 6 = 26. Watch out: solving for x = 20 and stopping is the classic trap; always re-read what's being asked.
Another way — level the tanks against the total:
If all four tanks matched tank 1, the total would be 4·(tank 1). The real total is 10 more (the built-in extras 0+1+3+6), so 4·(tank 1) = 90 − 10 = 80 → tank 1 = 20.
Jamar bought some pencils costing more than a penny each at the school bookstore and paid $1.43. Sharona bought some of the same pencils and paid $1.87. How many more pencils did Sharona buy than Jamar?
Show answer
Answer: C — 4 more pencils.
Show hints
Hint 1 of 2
Switch to whole cents (143 and 187) so you're working with integers. A whole number of pencils at a whole-cent price means that price divides each total exactly — so the price is a common factor of 143 and 187.
Still stuck? Show hint 2 →
Hint 2 of 2
This is a shared divisor hunt: the price divides both amounts, so it divides their gcd. Factor each amount and the price has to fall out (the "more than a penny" rules out the trivial 1¢).
Show solution
Approach: the price is a common factor of both totals
Some whole number of pencils times the (whole-cent) price gives 143¢, and likewise 187¢. So the price divides both 143 and 187 — it's a common factor.
Factor them: 143 = 11 × 13 and 187 = 11 × 17. Their only shared factor above 1 is 11, and "more than a penny" forbids 1¢, so each pencil costs 11¢.
The clean finish: Sharona paid 187 − 143 = 44¢ more, which is 44 / 11 = 4 extra pencils — no need to find each kid's count.
The transferable idea: "same whole-number price buys both totals" means the price is a common divisor; factoring or gcd pins it down. Subtracting the totals before dividing skips the individual counts.
2010 ends in 0, so it's begging to be split as 201 × 10. That instantly hands you the primes 2 and 5 — now just crack 201.
Still stuck? Show hint 2 →
Hint 2 of 2
To prime-factor, peel off the easy small primes first (2, 3, 5) using divisibility shortcuts, then factor whatever's left.
Show solution
Approach: peel off small primes using divisibility tests
The trailing 0 gives 2010 = 201 × 10 = 2 · 5 · 201. For 201, its digits sum to 3, so 3 divides it: 201 = 3 · 67.
67 has no small factor (not even, not a multiple of 3 or 5, and 7·7 > 67 once you've ruled out 7) — so it's prime.
Primes are 2, 3, 5, 67; sum = 77.
Why this transfers: divisibility tests (even → 2, digit-sum → 3, ends in 0/5 → 5) let you factor by inspection. And you only test primes up to the square root before declaring what's left prime.
For a MINIMUM, make each cube do double duty — one cube can show up in both the front view and the side view at once.
Still stuck? Show hint 2 →
Hint 2 of 2
Build the L-shaped front view with 3 cubes first, then add only what the side view still demands.
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Approach: reuse cubes across both views; add only what's missing
Minimizing means letting cubes count toward both pictures. Start with the front view, an L of 3 squares — that needs at least 3 cubes, sitting in one flat plane.
Now check the side view: it's also an L with depth, meaning something must sit behind the front row. Those 3 cubes alone give a side view only 1 cube deep, which is wrong. Adding one cube behind the corner fixes the side view — and that 4th cube touches the others, so the "every cube shares a face" rule holds.
No fewer than 4 can cover both an L front and an L side, so the minimum is 4 cubes.
You'll see this again: in "fewest cubes for these views" problems, the answer is driven by where the two views disagree — build the bigger view, then patch only the parts the other view still forces.
Don't add 1996 terms one at a time. Look at the signs: +, −, −, +, then they repeat. A repeating sign pattern begs you to chop the sum into matching chunks. Try grouping four terms at a time.
Still stuck? Show hint 2 →
Hint 2 of 2
Group as (1 − 2 − 3 + 4) + (5 − 6 − 7 + 8) + …. Compute just ONE block; if every block gives the same thing, you only need to know how many blocks there are.
Show solution
Approach: group into blocks of four
The signs cycle +, −, −, + with period 4, so split into blocks of four consecutive numbers. The first block is 1 − 2 − 3 + 4 = 0, and every later block has the same shape (e.g. 5 − 6 − 7 + 8 = 0), so each one is 0 too.
Since 1996 = 4 × 499, the sum is exactly 499 blocks, each worth 0 — total 0. (No need to check leftovers: 1996 divides evenly by 4.)
Why this transfers: a repeating sign or value pattern means 'group by the period, evaluate one group, multiply by the count.' Always check whether the length is a clean multiple of the period — leftovers are where these problems hide their answer.
The two schools are different sizes, so you can't just average 11% and 17%. A percent of 100 students and a percent of 200 students mean different numbers of kids — convert to actual kids first.
Still stuck? Show hint 2 →
Hint 2 of 2
Turn each grade-6 percent into a head count, add the counts, then take that over the 300 total.
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Approach: convert to head counts (the schools have different sizes), then one combined percent
The trap is averaging 11% and 17% to get 14%. That's wrong because Cleona has twice as many students — its percentage should count double. So switch to real counts.
Grade 6 counts: Annville 11% of 100 = 11 kids; Cleona 17% of 200 = 34 kids. Together 45 kids.
Out of all 300 students that's 45300 = 15%.
Why this transfers: this is a weighted average — percentages can only be combined directly when the groups are the same size. Different sizes → count the actual items, then re-percent at the end. (Notice 15% leans toward Cleona's 17%, because Cleona is the bigger school.)
'Fewest posts' is a hint about which side to hide against the wall — the wall side needs no fence and no posts of its own, so put the *longest* side there to fence the least length.
Still stuck? Show hint 2 →
Hint 2 of 3
Now you have one straight path of fence (unrolled, it's just a line). The classic trap: a 132 m line with posts every 12 m has 132⁄12 = 11 gaps but 12 posts — one more post than gaps, because both ends get a post.
Still stuck? Show hint 3 →
Hint 3 of 3
Check the corners: at 36 m and 96 m the fence bends, but both are multiples of 12, so a post already lands there — no bonus posts needed.
To use the *fewest* posts, fence the *least* length, so press the longest side (60 m) against the wall — that side needs no fence. The remaining three sides total 36 + 60 + 36 = 132 m of fence.
Think of that 132 m as one straight run (the corners don't change its length). Posts sit every 12 m, so there are 132⁄12 = 11 gaps — but the number of posts is one more than the gaps because both endpoints get a post: 11 + 1 = 12.
Double-check the bends at 36 m and 96 m: both are multiples of 12, so posts already land exactly on the corners — nothing extra to add.
Why 'posts = gaps + 1': this is the fencepost principle — the off-by-one that bites people who just divide and forget the two endpoints. Cutting a line into n equal pieces always needs n + 1 marks.
The sum of 25 consecutive even integers is 10,000. What is the largest of these 25 consecutive integers?
Show answer
Answer: E — 424.
Show hints
Hint 1 of 2
Evenly-spaced numbers are symmetric: every term below the center is balanced by an equal-sized term above it. So the AVERAGE of the whole list sits exactly on the MIDDLE term — you never have to write out all 25 numbers.
Still stuck? Show hint 2 →
Hint 2 of 2
Get the middle term as total ÷ count, then climb from the center to the largest: the top term is 12 steps of 2 above the middle.
Show solution
Approach: the middle term IS the average; then step out to the largest
For 25 evenly-spaced numbers, the average equals the middle (13th) term: 10,000 ÷ 25 = 400.
From the 13th term up to the 25th is 12 steps, each of size 2, so the largest = 400 + 12 × 2 = 424.
Why this transfers: for any arithmetic sequence with an ODD number of terms, average = middle term — turning a sum problem into a one-line lookup. (With an even count, the average lands halfway between the two middle terms.)
In a certain year, January had exactly four Tuesdays and four Saturdays. On what day did January 1 fall that year?
Show answer
Answer: C — Wednesday.
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Hint 1 of 3
January has 31 days, and 31 = 4 weeks + 3 leftover days. That means SOME weekdays squeeze in a 5th time. Which ones? The first three days of the month — they're the extras.
Still stuck? Show hint 2 →
Hint 2 of 3
So exactly three weekdays appear 5 times (the day Jan 1 lands on, plus the next two), and the other four appear only 4 times. The problem wants Tuesday AND Saturday to be among the rare (4-time) ones.
Still stuck? Show hint 3 →
Hint 3 of 3
For both Tuesday and Saturday to appear only 4 times, neither can be in the 'first three days' trio. Test each possible starting day and see whose trio of three consecutive weekdays misses both Tue and Sat.
Show solution
Approach: find which start makes Tue and Sat both rare
31 days = 4 full weeks (28 days) + 3 extra days. Every weekday shows up 4 times for the 28, and the 3 extras — Jan 1, 2, 3's weekdays — get a 5th appearance. So three consecutive weekdays appear 5 times; the rest appear 4.
We need Tuesday and Saturday to both be 4-time days, i.e. NOT in that trio of three-in-a-row starting at Jan 1. List the trios: a Monday start gives Mon-Tue-Wed (hits Tue ✗); a Tuesday start hits Tue ✗; a Wednesday start gives Wed-Thu-Fri — misses both Tue and Sat ✓.
So January 1 fell on Wednesday.
Why this transfers: in any month, take the day count mod 7 to find how many 'extra' days there are; those extras (counting from the 1st) are exactly the weekdays that occur one more time than the others.
A list of 8 numbers is formed by beginning with two given numbers. Each new number in the list is the product of the two previous numbers. Find the first number if the last three numbers are 16, 64, 1024.
Show answer
Answer: B — 1/4.
Show hints
Hint 1 of 2
You're given the END of the list and asked for the START — that's a signal to run the rule *backward*. The forward rule is 'multiply the two before,' so the reverse is a division. Which two numbers do you divide?
Still stuck? Show hint 2 →
Hint 2 of 2
If a term equals (term before it) × (term two before it), then 'two before' = (this term) ÷ (the one just before). Keep peeling backward one step at a time until you reach the first number.
Show solution
Approach: work backward — undo each multiply with a divide
Label the list t₁…t₈; the rule is tₙ = tₙ₋₁ × tₙ₋₂. We know t₆=16, t₇=64, t₈=1024 (and indeed 16×64 = 1024 ✓, a good consistency check).
So the first number is 1/4. (Verify forward: 1/4, 4 → 1, 4, 4, 16, 64, 1024 ✓.)
*Why this transfers:* when a problem hands you the end of a chain, reverse the operation (multiply→divide, add→subtract) and walk back — always sanity-check by running it forward.
The hundreds digit of a three-digit number is 2 more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?
Show answer
Answer: E — 8.
Show hints
Hint 1 of 3
When you reverse a 3-digit number, the tens digit doesn't move — only the hundreds and units swap. So in the subtraction, the tens completely cancel and won't affect the answer.
Still stuck? Show hint 2 →
Hint 2 of 3
Reversing always gives a difference of 99 × (hundreds digit − units digit). Here that gap is fixed at 2, so the difference is a single fixed number — no variables survive.
Still stuck? Show hint 3 →
Hint 3 of 3
So you don't need the actual digits: just compute 99 × 2 and read off its units digit.
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Approach: the tens digit cancels; difference is forced
Write the number as 100·(hundreds) + 10·(tens) + (units). Reversing swaps hundreds and units, so when you subtract, the 10·(tens) terms cancel exactly.
What's left is 99·(hundreds − units). The hundreds digit is 2 more than the units, so hundreds − units = 2, every time.
Difference = 99 × 2 = 198, whose units digit is 8.
Why this transfers: ‘a number minus its reversal’ is always a multiple of 99 (for 3 digits), and only the gap between the outer digits matters. Spotting that the middle digit cancels means you can answer without ever choosing specific digits.
Another way — try a concrete example:
Pick any number fitting the rule, say hundreds 2 more than units: 301. Reverse to 103.
301 − 103 = 198 ⇒ units digit 8. Trying 412 − 214 = 198 too — the units digit is locked at 8 no matter what you choose, which is the whole point.
An equilateral triangle and a regular hexagon have equal perimeters. If the triangle's area is 4, what is the area of the hexagon?
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Answer: C — Area 6.
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Hint 1 of 2
A regular hexagon is secretly 6 little equilateral triangles meeting at its center — so the whole problem is about comparing those small triangles to the big one. First find the hexagon's side from the equal-perimeter clue.
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Hint 2 of 2
Equal perimeters: the triangle's 3 sides equal the hexagon's 6 sides, so each hexagon side is half the triangle's side. Key fact: when you halve a length, area shrinks by the square — to 1/4, not 1/2.
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Approach: hexagon = six equilateral triangles; area scales as (length)²
Equal perimeters means 3 × (triangle side) = 6 × (hexagon side), so the hexagon's side is half the triangle's side.
Cut the hexagon into its 6 natural equilateral triangles (from center to the corners) — each has that half-length side.
Halving the side scales area by (1/2)² = 1/4 (area always scales as the square of length), so each mini-triangle has area 4 × 1/4 = 1.
Six of them: hexagon area = 6 × 1 = 6.
The big takeaway: area scales like the square of the length ratio — double the side, 4× the area; half the side, 1/4 the area. This single fact handles most "similar figures" comparisons without any formula for the area itself.
The dip between the peaks is exactly where the two mountains OVERLAP. So if you treat each mountain as a full triangle and add them, you've counted that dip twice — the setup for inclusion–exclusion.
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Hint 2 of 2
Handy fact: a 45-45-90 mountain triangle of peak height H has base 2H, so its area is 12(2H)(H) = H2. The dip is a third such triangle of height h.
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Approach: inclusion–exclusion on three 45-45-90 triangles
First, a clean area shortcut: each mountain is a right-isoceles triangle (90° peak, 45° base angles), so its base is 2×(its height) and its area is 12(2H)(H) = H2. Heights 8 and 12 give areas 64 and 144.
If you just add 64 + 144, you double-count the V-dip where the two mountains overlap — and that dip is itself a 45-45-90 triangle of height h, area h2. So the true artwork area is 64 + 144 − h2.
Set that to the given 183: 208 − h2 = 183 → h2 = 25 → h = 5. This transfers: when two regions overlap, area(A) + area(B) − area(overlap) = total — inclusion–exclusion turns a tricky shape into three simple ones.
Two 4 × 4 squares intersect at right angles, bisecting their intersecting sides, as shown. The circle's diameter is the segment between the two points of intersection. What is the area of the shaded region created by removing the circle from the squares?
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Answer: D — 28 − 2π.
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Hint 1 of 3
The shaded area is 'both squares, minus the circle'. The trap is double-counting the overlap, so first nail the combined area of the two squares using inclusion–exclusion: square + square − the piece they share.
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Hint 2 of 3
The shared piece is the small square where they cross. 'Bisecting their intersecting sides' means each square is cut at the midpoints of those sides, so the overlap is a 2×2 square (area 4).
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Hint 3 of 3
For the circle, the diameter is the segment joining the two crossing points — that's the diagonal of the 2×2 overlap square. Diagonal = 2√2, so radius = √2.
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Approach: inclusion–exclusion, then subtract the circle
Combined area of the two squares (avoid double-counting the middle): 16 + 16 − (overlap). The overlap is a 2×2 square because the squares meet at the midpoints of the crossed sides, so overlap = 4. Combined region = 16 + 16 − 4 = 28.
Circle: its diameter is the diagonal of that 2×2 overlap square = 2√2, so radius = √2 and area = π(√2)² = 2π.
Shaded = combined region − circle = 28 − 2π.
Why these two ideas pair up so often: inclusion–exclusion (add the parts, subtract the shared piece once) handles the overlapping squares, and the √2 comes from a square's diagonal — both are workhorses you'll reuse constantly. Estimate check: 2π ≈ 6.3, so the shaded area is about 28 − 6.3 ≈ 21.7, comfortably less than the 28 of squares alone.
