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Problem 1 · 2009 AMC 8
Easy
Algebra & Patternswork-backward
Bridget bought a bag of apples at the grocery store. She gave half of the apples to Ann. Then she gave Cassie 3 apples, keeping 4 apples for herself. How many apples did Bridget buy?
Show answer
Answer: E — 14 apples.
Show hints
Hint 1 of 2
The story runs forward (give away, give away, keep), but you know the END, not the start. So play the movie in reverse.
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Hint 2 of 2
To undo a story, reverse the order AND flip each action: gave away → add back, took half → double.
Show solution
Approach: work backward, inverting each step
Start from what's known: 4 apples kept. The last thing before that was giving Cassie 3, so undo it — add them back: 4 + 3 = 7. That 7 is what she had right after giving Ann half.
Giving away half means 7 is the OTHER half, so the original was double: 2 × 7 = 14.
Why this transfers: whenever a problem hands you the finish and asks for the start, work-backward turns it into pure arithmetic — just remember to invert each operation (subtract becomes add, halve becomes double).
Another way — algebra (one equation):
Let she bought n. After giving Ann half she has n/2; after Cassie's 3 she has n/2 − 3 = 4.
Don't add left to right. In an evenly-spaced run, pair the smallest with the largest: 90 + 99, 91 + 98, 92 + 97… each pair makes the same total. What is that total, and how many pairs are there?
Still stuck? Show hint 2 →
Hint 2 of 2
Sum of an evenly-spaced run = (first + last)⁄2 × count — the (first+last)⁄2 is just the average, sitting right in the middle. So a long sum becomes one multiplication.
Show solution
Approach: average × count
These ten numbers climb by 1, so their average is the middle value: (90 + 99)⁄2 = 94.5. There are 10 of them.
Sum = 94.5 × 10 = 945 — the average times the count.
Sanity check: all ten numbers are in the 90s, so the total must be between 10×90 = 900 and 10×99 = 990. 945 lands right in the middle, as it should.
Another way — pair from the ends:
Pair 90+99, 91+98, 92+97, 93+96, 94+95 — five pairs, each summing to 189.
Don't add ten numbers in a row — notice they're evenly spaced. What does the smallest plus the largest equal, and does that hold for the next pair in?
Still stuck? Show hint 2 →
Hint 2 of 2
In any evenly-spaced list, pairing the ends inward gives equal sums. Count the pairs.
Show solution
Approach: pair from the ends
The ten numbers climb by 2 each time, so pairing ends inward gives a constant total: 81 + 99 = 180, 83 + 97 = 180, and so on. Ten numbers make five pairs, each 180, so the bracket is 5 × 180 = 900.
Then double: 2 × 900 = 1800.
Why this transfers: this is the Gauss trick — any evenly-spaced sum equals (count) × (average of first and last). Here that's 10 × (81+99)/2 = 10 × 90 = 900, the same 900 without listing pairs.
The shape has one corner pushed IN (the bite near E and F). Instead of chopping it into pieces, imagine filling that bite to make a full rectangle — then take the bite back out. What is the full rectangle?
Still stuck? Show hint 2 →
Hint 2 of 2
This is the bounding-box trick: enclose any rectilinear shape in the smallest rectangle that contains it, then subtract the rectangular pieces that aren't part of the shape. One subtraction beats slicing into several rectangles.
Show solution
Approach: bounding rectangle minus the notch
Fill the bite to complete the full rectangle: it is 6 wide (the top AB) by 9 tall (the right side BC), so its area is 6 × 9 = 54.
The bite that was removed is itself a small rectangle. Its width is 6 − 4 = 2 (top width minus bottom width DC) and its height is 9 − 5 = 4 (right side minus left side AF), so its area is 2 × 4 = 8.
Polygon = full rectangle − bite = 54 − 8 = 46.
Why this transfers: any L-shape, T-shape, or staircase is just a rectangle with rectangular bites taken out — find the missing side lengths by subtracting the parts you know, then subtract areas.
Another way — split into two rectangles:
Cut horizontally at the level of F/E. Top piece: 6 wide × 5 tall = 30. Bottom piece: 4 wide × (9 − 5) = 4 × 4 = 16.
30 + 16 = 46 — same answer, built up instead of cut down.
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)
Show hints
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 \(12\).
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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.
Each square is built on a side of the triangle, so the side of a square IS a side of the triangle — square-root each area.
Still stuck? Show hint 2 →
Hint 2 of 2
5, 12, 13 is a famous Pythagorean triple, so the triangle has a right angle and the two shorter sides are its legs.
Show solution
Approach: areas give the side lengths; recognize the right triangle
A square's side length is the square root of its area, and each square shares a side with the triangle. So the triangle's sides are √169 = 13, √144 = 12, and √25 = 5.
Check 5² + 12² = 25 + 144 = 169 = 13²: the sides fit the Pythagorean rule, so the angle between the two short sides is a right angle. That makes 5 and 12 the perpendicular legs — the base and height.
Area of a right triangle = ½ × leg × leg = ½ × 5 × 12 = 30.
Worth keeping: 5-12-13 and 3-4-5 (and its scalings) are the triples to memorize — spotting one instantly tells you a triangle is right-angled, no slow Pythagorean computation needed.
The rectangle is tiny and way off to the right (x only runs 15 to 16). So instead of graphing the whole lines, just ask: when a line reaches that far right, is its height between 3 and 5? You only need to test x = 15 and x = 16.
Still stuck? Show hint 2 →
Hint 2 of 2
Read each line's slope off its two points to get its equation. Line AB rises 1 for every 3 right: y = x/3. Line CD drops 1 for every 2 right: y = 10 − x/2. Plug in x = 15 and 16.
Show solution
Approach: evaluate the two lines at the rectangle's x-range
Notice the rectangle is a thin sliver far to the right: x runs only 15 to 16, y only 3 to 5. So you don't graph the whole lines — you just check whether either line is at the right height when it gets out to x = 15 or 16.
Line AB: slope = 1/3 (from A(0,0) to B(3,1)), so y = x/3. At x = 15: y = 5 — exactly the corner (15, 5) ✓. At x = 16: y ≈ 5.33, just above the box.
Line CD: slope = −1/2, so y = 10 − x/2. At x = 15: y = 2.5; at x = 16: y = 2 — both below the box.
Only (15, 5) lands on the rectangle — 1 point. Why this is faster: when a region is far from the lines, test the region's edges rather than tracing the lines all the way out.
The table shows some of the results of a survey by radio station KACL. What percentage of the males surveyed listen to the station? (Total surveyed: 200. Females: 96. Females who listen: 58. Males who don't listen: 26. Total listeners: 136. Total non-listeners: 64.)
Show answer
Answer: E — 75%.
Show hints
Hint 1 of 2
A two-way table is built on the rule that rows and columns must add to their totals. Find a cell you CAN fill (males total), then chase the rest.
Still stuck? Show hint 2 →
Hint 2 of 2
You only need the male row: total males, then split into listen/don't-listen. Ignore the female numbers once you have the male total.
Show solution
Approach: complete the male row, then take the percentage
Total males = total surveyed − total females = 200 − 96 = 104.
Males who listen = males total − males who don't = 104 − 26 = 78.
Percentage of males who listen = 78 ÷ 104 = 75%.
Shortcut worth seeing: instead of finding 78, note 26 out of 104 is exactly ¼ (since 26 × 4 = 104), so the non-listeners are 25% and the listeners are 100% − 25% = 75%. Spotting the clean fraction beats the division.
A fruit bowl has 3 apples, 2 oranges, and 4 bananas. Judy will take at least one piece of fruit. How many different fruit combinations can she take? (Pieces of the same fruit look alike, so only how many of each she takes matters.)
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Answer: 59 combinations
Show hints
Hint 1 of 4
Since pieces of the same fruit are identical, for each fruit only the quantity matters — like the porter's coins.
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Hint 2 of 4
For each fruit she may take 0 up to all of them. Count the quantity choices for each fruit.
What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?
Show answer
Answer: C — 330.
Show hints
Hint 1 of 2
Don't compute the huge LCM and tiny GCD separately and then divide. Once both numbers are in prime-factored form, the ratio reads off prime by prime: for each prime, LCM takes the bigger exponent and GCD the smaller, so the ratio just keeps the difference.
Still stuck? Show hint 2 →
Hint 2 of 2
Prime factorization is the master key for LCM and GCD: GCD = product of min exponents, LCM = product of max exponents. Compare the two prime towers side by side.
Show solution
Approach: prime factor each number, then compare exponents
Factor: 180 = 22 · 32 · 5 and 594 = 2 · 33 · 11.
GCD takes the smaller power of each shared prime: 21 · 32 = 18.
LCM takes the larger power of every prime that appears: 22 · 33 · 5 · 11 = 4 · 27 · 55 = 5940.
Ratio: 5940 ÷ 18 = 330.
Another way — ratio straight from exponent gaps (no big numbers):
LCM/GCD keeps, for each prime, the gap between the two exponents: 2 has exponents 2 and 1 (gap 1), 3 has 2 and 3 (gap 1), and 5, 11 appear in only one number (full power).
So the ratio = 21 · 31 · 5 · 11 = 2 · 3 · 5 · 11 = 330 — no 5940 or 106,920 ever needed.
Another way — product identity LCM · GCD = a · b:
Use LCM(a,b) · GCD(a,b) = a · b, so LCM/GCD = ab/GCD2.
"What percent preferred blue" is blue out of EVERYONE — not blue out of the tallest bar. What's the denominator you actually need?
Still stuck? Show hint 2 →
Hint 2 of 3
A percent always needs its whole: percent = part ÷ total. Here the total is every bar added together, so read them all before dividing.
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Hint 3 of 3
Line up the bar tops with the frequency scale carefully — each gridline is worth 20.
Show solution
Approach: blue's count divided by the grand total of all bars
Read each bar against the scale: Red 50, Blue 60, Brown 40, Pink 60, Green 40. The total surveyed is 50 + 60 + 40 + 60 + 40 = 250.
Blue's share is 60 out of 250: 60 ÷ 250 = 0.24 = 24%.
Why this transfers: the most common percent-from-a-graph mistake is forgetting to total ALL the bars — the part means nothing without its whole. Always build the denominator first.
Sanity check: blue is one of five bars, so a fair share would be 20%. Blue is a little above average height, so a little above 20% — 24% fits.
A circle's reach is capped by the closest bit of boundary. On this lumpy plus-shape, that closest bit isn't a flat edge — it's an inward corner where two arms meet.
Still stuck? Show hint 2 →
Hint 2 of 2
Center the circle in the middle (by symmetry). Find one nearest inward corner: it's 1 unit across and 2 units up from the center. The Pythagorean theorem gives the radius — then square it for the area.
Show solution
Approach: radius = distance to the closest point of the boundary
Center the circle at the region's center (forced by symmetry). The biggest it can grow is limited by the nearest boundary point — and on this cross-shape those are the inward corners where the arms notch in, not the outer flat edges.
One such corner sits 1 unit across and 2 units up from the center, so the radius is the hypotenuse: √(12 + 22) = √5.
Area = π × (√5)2 = 5π.
Why this transfers: the largest inscribed circle is always pinned by the closest point of the boundary — so on a non-convex shape, hunt the inward corners first. And notice you never needed √5 itself: squaring it for the area gives a clean 5.
A drawer holds \(6\) red, \(7\) green, \(4\) blue, and \(9\) yellow ribbons, all the same shape so you can't tell them apart by feel. In the dark, how many ribbons must you pull out to be sure you have at least one of EVERY color?
Show answer
Answer: 23 ribbons
Show hints
Hint 1 of 3
Ask the key question: with the worst luck, how many ribbons could you pull and still be missing one color (have only \(3\) of the \(4\) colors)?
Still stuck? Show hint 2 →
Hint 2 of 3
To stay missing a color as long as possible, imagine grabbing every ribbon of the three biggest colors first, and none of the smallest.
Still stuck? Show hint 3 →
Hint 3 of 3
The smallest color is blue (\(4\)). Add up the other three counts: \(6 + 7 + 9\). After all those, you still have no blue. What does the next ribbon have to be?
Show solution
Approach: Asking the key question — the worst-case grab
Ask the key question: what is the most ribbons you can hold while still missing a color?
The unluckiest run grabs every ribbon of the three most common colors and skips the rarest (blue): \(6\) (red) \(+\ 7\) (green) \(+\ 9\) (yellow) \(= 22\) ribbons, and you still have zero blue.
The \(23\)rd ribbon has no choice but to be blue, completing all four colors. So you must take \(23\) ribbons to be certain. (You might get lucky much sooner, but 'certain' means even in the worst case.)
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.
Still stuck? Show hint 2 →
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.
Show solution
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.
How many whole numbers between 100 and 400 contain the digit 2?
Show answer
Answer: C — 138.
Show hints
Hint 1 of 2
Counting numbers that contain a 2 directly is messy — a 2 might be in the hundreds, tens, or units place, and numbers like 220 or 282 would get counted twice. When 'at least one' is hard, count the OPPOSITE.
Still stuck? Show hint 2 →
Hint 2 of 2
Count how many numbers have NO 2 at all, then subtract from the total. 'Total − (none) = (at least one)' sidesteps all the double-counting. The numbers run 100 to 399, so there are 300 of them.
Show solution
Approach: complement count
There are 300 whole numbers from 100 to 399. Instead of counting those that contain a 2, count those that contain NO 2 (much cleaner — no overlaps).
No-2 count: the hundreds digit can be 1 or 3 (2 ways, since 2 is banned and a 3-digit number can't start with 0), and each of the tens and units can be any digit except 2 (9 ways each): 2 × 9 × 9 = 162.
Contains a 2 = total − none = 300 − 162 = 138.
Why this transfers: whenever you want 'at least one' of something, counting the cases with NONE and subtracting is almost always easier — it turns an overlapping mess into one clean multiplication.
Joyce made 12 of her first 30 shots in the first three games of this basketball game, so her seasonal shooting average was 40%. In her next game, she took 10 shots and raised her seasonal shooting average to 50%. How many of these 10 shots did she make?
Show answer
Answer: E — 8.
Show hints
Hint 1 of 2
A season average is about totals, not single games. Work out her total makes before and after the new game, and the difference is what you want.
Still stuck? Show hint 2 →
Hint 2 of 2
After the extra game she's taken 30 + 10 = 40 shots at a 50% average. How many total baskets does that mean — and how many had she already made?
Show solution
Approach: work in season totals, then take the difference
An average is (total made) ⁄ (total shots), so work with totals. After the game she's taken 40 shots at 50%, meaning 50% × 40 = 20 total makes.
She'd already made 12, so this game she made 20 − 12 = 8.
Why this transfers: never average the averages or work game-by-game — convert each percentage back to a count of makes over a count of shots, then subtract. Totals are what actually add up.
Two trains are \(200\) miles apart on the same track, heading toward each other. One goes \(60\) mph, the other \(40\) mph. A fly starts on the front of the slower train and flies back and forth between the two trains at \(240\) mph, turning around instantly each time it reaches a train, until the trains crash and squash it. How far does the fly travel in total?
Show answer
Answer: 480 miles
Show hints
Hint 1 of 3
Adding up the fly's endless shrinking back-and-forth trips is a nightmare. Ask the key question: what single thing, times the fly's speed, gives its total distance?
Still stuck? Show hint 2 →
Hint 2 of 3
Distance = speed \(\times\) time, and the fly is flying the WHOLE time until the trains crash. So the real question is: how long until they crash?
Still stuck? Show hint 3 →
Hint 3 of 3
The two trains close the \(200\)-mile gap together at \(60 + 40 = 100\) mph. Find how long that takes, then multiply by the fly's \(240\) mph.
Show solution
Approach: Asking the key question — find the time, not the zig-zags
Don't add up the fly's zig-zags. Ask the key question: how long does the fly fly? Its distance is just speed times time, and it flies until the trains crash.
The trains approach each other at a combined speed of \(60 + 40 = 100\) mph. To close a \(200\)-mile gap at \(100\) mph takes time \(= \frac{200}{100} = 2\) hours.
The fly flies for those whole \(2\) hours at \(240\) mph: \(240 \times 2 = 480\) miles.
Gage skated 1 hr 15 min each day for 5 days and 1 hr 30 min each day for 3 days. How long would he have to skate the ninth day in order to average 85 minutes of skating each day for the entire time?
Show answer
Answer: E — 2 hours.
Show hints
Hint 1 of 2
An average is a target everyone must *balance out* to. So measure each day as how far above or below 85 it lands, not its raw minutes.
Still stuck? Show hint 2 →
Hint 2 of 2
Days below 85 owe minutes; days above 85 pay them back. Net the surplus against the shortfall — day 9 must cover whatever's still owed (on top of its own 85).
Show solution
Approach: measure each day against the 85-minute target
Read every day as a *deviation* from the 85-min target. The 5 days at 75 are each 10 short → −50; the 3 days at 90 are each 5 over → +15.
Running balance: −50 + 15 = −35, a 35-minute shortfall. Day 9 must hit the target *and* erase it: 85 + 35 = 120 min = 2 hours.
*The transferable trick:* to hit a target average, track only the ± gaps from the target — they must cancel to zero. The last value just absorbs the leftover gap, far lighter than summing everything.
Another way — totals:
Skated so far: 5·75 + 3·90 = 645 min. Needed for the average: 9·85 = 765 min.
Guess where they meet first: the middle section, where both happen to drive 40 mph. There, equal speeds mean they close the gap evenly — all the trickiness is in how unevenly they reach the middle.
Still stuck? Show hint 2 →
Hint 2 of 2
Time A to the middle: 5/25 = 1/5 hr. Time B to the middle: 5/20 = 1/4 hr. So A enters the middle 1/20 hr early — turn that head start into miles, then split what's left.
Show solution
Approach: find each car's position when both are in the equal-speed middle section
Bet that they meet in the middle 40-mph section (then they go the same speed, so you only need positions). A covers its 5-mile left section in 5/25 = 1/5 hr; B covers its 5-mile right section in 5/20 = 1/4 hr. So A reaches the middle 1/4 − 1/5 = 1/20 hr before B.
In that 1/20-hr head start, A drives 40 × 1/20 = 2 miles into the middle. So when B finally enters, A is at mile 7, B at mile 10 — a 3-mile gap, both now at 40 mph.
Equal speeds split the gap evenly: each drives 1.5 more miles. A's meeting point is 7 + 1.5 = 8.5 miles from A.
Why this transfers: when speeds match over the stretch where they meet, the closing is symmetric — so isolate the asymmetry (the unequal times to arrive), convert it to a distance, then share the remainder equally.
Another way — track total time to the meeting instant:
They meet somewhere in the middle section at the same clock time t. Suppose A has gone a distance d into the middle and B has gone 5 − d into it (their middle distances fill the 5-mile section).
Let PQRS be a square piece of paper. P is folded onto R, and then Q is folded onto S. The area of the resulting figure is 9 square inches. Find the perimeter of square PQRS.
Show answer
Answer: D — 24 inches.
Show hints
Hint 1 of 2
Don't picture the exact shape after folding — track only the AREA. Each fold lays the paper onto itself, so the visible area halves every time.
Still stuck? Show hint 2 →
Hint 2 of 2
Two folds means the final area is the whole square's area cut in half twice — that's one quarter. Set that quarter equal to 9 and work back to the side.
Show solution
Approach: each fold halves the area; two folds leave a quarter
Folding a flat sheet onto itself doubles its thickness and halves its visible area. Fold P onto R: area halves. Fold Q onto S: it halves again. So the final figure is ¼ of the square.
That quarter is 9 in²: ¼·s² = 9, so the full square has area s² = 36, and its side is s = 6.
Perimeter = 4 × side = 4 × 6 = 24 inches.
Why this transfers: when a problem folds paper but only asks about area, skip the geometry of the resulting shape — just multiply the original area by ½ for each fold. Then a quick √ takes area back to side length.
Steph scored 15 baskets out of 20 attempts in the first half of a game, and 10 baskets out of 10 attempts in the second half. Candace took 12 attempts in the first half and 18 attempts in the second. In each half, Steph scored a higher percentage of baskets than Candace. Surprisingly they ended with the same overall percentage of baskets scored. How many more baskets did Candace score in the second half than in the first?
Show answer
Answer: C — 9 more baskets.
Show hints
Hint 1 of 2
Both players took the same number of total attempts (30). If two people with equal attempts finish at the same overall percentage, what must be equal about their makes? That collapses the “surprising” clue into a hard number.
Still stuck? Show hint 2 →
Hint 2 of 2
Candace also made 25 baskets. Now her per-half percentages are each strictly below Steph's, which caps her first-half and second-half makes — and only one split of 25 fits both caps.
Show solution
Approach: equal attempts + equal overall % forces equal total makes; then squeeze the split
Insight: turn the “surprising” tie into arithmetic. Both shot 30 total (Steph 20+10, Candace 12+18). Equal attempts and equal overall percentage means equal makes — Steph made 15 + 10 = 25, so Candace made 25 too.
Let Candace's makes be f (of 12) and s (of 18), with f + s = 25. Beating-by-Steph in each half caps her: f/12 < 15/20 = ¾ forces f ≤ 8, and s/18 < 1 forces s ≤ 17.
Those caps add to exactly 8 + 17 = 25, so the only split is f = 8, s = 17 — any less in one half can't be made up in the other.
s − f = 17 − 8 = 9.
Why the caps pin it down: when two upper bounds sum to exactly the required total, each variable is pinned to its max — no slack to trade. (This is also the resolution of the classic “Simpson's paradox” setup: losing both halves yet tying overall.)
The number 6545 can be written as a product of a pair of positive two-digit numbers. What is the sum of this pair of numbers?
Show answer
Answer: A — 162.
Show hints
Hint 1 of 2
You can't see the two-digit factors by staring at 6545 — but its prime factorization reveals every possible way to split it. Break it into primes first; the answer is just a regrouping of those.
Still stuck? Show hint 2 →
Hint 2 of 2
6545 ends in 5, so 5 comes out immediately. Keep pulling small primes, then sort the prime pieces into two groups that each land between 10 and 99.
Show solution
Approach: prime-factor, then regroup the primes into two two-digit numbers
Break 6545 down: it ends in 5 → divide by 5 to get 1309; 1309 = 7 × 187 = 7 × 11 × 17. So 6545 = 5 × 7 × 11 × 17.
Now bundle those four primes into two two-digit numbers. The bundle (5 × 17) × (7 × 11) = 85 × 77 works — both are two-digit. (5×7=35 with 11×17=187 fails, since 187 is three digits.)
Sum = 85 + 77 = 162.
Why this transfers: the prime factorization is the master list of a number's building blocks — every factor pair is just one way of dividing those primes into two teams. Factor first, then the regrouping is quick.
The 'mediant' of two fractions adds the tops and adds the bottoms: \(\frac{a}{b} \oplus \frac{c}{d} = \frac{a+c}{b+d}\). (This is NOT how you add fractions, but the result always lands strictly between them.) What is the mediant of \(\frac{1}{3}\) and \(\frac{1}{2}\)? Give it as a fraction in lowest terms.
Show answer
Answer: 2/5
Show hints
Hint 1 of 4
Compute the mediant of \(\frac{1}{3}\) and \(\frac{1}{2}\) by adding tops and adding bottoms.
Still stuck? Show hint 2 →
Hint 2 of 4
Turn all three fractions into decimals (or a common denominator) and put them in order. Is the mediant in the middle?
Still stuck? Show hint 3 →
Hint 3 of 4
To see why it always works, compare \(\frac{a}{b}\) with \(\frac{a+c}{b+d}\) using cross-multiplication (the bigger fraction has the bigger cross-product).
Show solution
Approach: Compute the mediant, verify it lies between via cross-multiplication
Add tops and bottoms: \(\frac{1 + 1}{3 + 2} = \frac{2}{5}\), already in lowest terms.
Check the order as decimals: \(\frac{1}{3} \approx 0.333\), \(\frac{2}{5} = 0.4\), \(\frac{1}{2} = 0.5\). The mediant sits right between them.
Why it always works: if \(\frac{a}{b} < \frac{c}{d}\) then \(ad < bc\). Comparing \(\frac{a}{b}\) with \(\frac{a+c}{b+d}\) by cross-multiplying gives \(a(b+d) < b(a+c)\), i.e. \(ab + ad < ab + bc\), i.e. \(ad < bc\) — exactly what we know.
The same check shows the mediant is below \(\frac{c}{d}\), so the mediant of \(\frac{1}{3}\) and \(\frac{1}{2}\) is \(\frac{2}{5}\) and always lands strictly between.
Pick a friendly diameter so the 2:3 split is whole: AE = 10 gives AC = 4, CE = 6, big radius 5. Then build the wavy upper region out of clean half-disks rather than fighting the S-curve directly.
Still stuck? Show hint 2 →
Hint 2 of 2
Decompose a curvy region into known semicircles: start with the big upper half-disk, then subtract the bump that dips below the dividing line and add the bump that rises above it.
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Approach: big half-disk, minus the dipping bump, plus the rising bump
Let AE = 10, so AC = 4 (radius 2) and CE = 6 (radius 3); the big circle has radius 5. Half-areas: big = ½·π·5² = 12.5π, small ABC = ½·π·2² = 2π, small CDE = ½·π·3² = 4.5π.
The S-shaped boundary carves the upper region as: big upper half-disk − semicircle ABC (it pushes down, removing area) + semicircle CDE (it bulges up, adding area) = 12.5π − 2π + 4.5π = 15π.
The whole circle is 25π, so the lower region is 25π − 15π = 10π. Ratio = 15π : 10π = 3 : 2.
Sanity check: the upper region is the bigger one, matching the shaded (mostly-top) figure. Notice: the π's cancel in the ratio — area-ratio problems rarely need the actual value of π.
Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers? (Visible sides: 44, 59, 38.)
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Answer: B — 14.
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Hint 1 of 3
Look at the three visible numbers: 44 and 38 are even, but 59 is odd. Since all three card-sums are EQUAL, that odd one out is the lever — what kind of prime must hide behind it?
Still stuck? Show hint 2 →
Hint 2 of 3
Almost every prime is odd; the only even prime is 2. Use parity: even + (odd prime) is odd, while odd + (odd prime) is even — so the sums can only all match if one special card hides the 2.
Still stuck? Show hint 3 →
Hint 3 of 3
Once parity pins down which card hides 2, you get the common sum, then subtract each visible number to recover the other two hidden primes — and check they're actually prime.
Show solution
Approach: parity of the visible numbers forces where 2 hides
The three sums are equal, so they're all the same parity (all odd or all even). Behind 44 and 38 (even), an odd prime gives an odd sum; behind 59 (odd), an odd prime gives an even sum. Those can't match — so something must break the "odd prime" assumption.
The only escape is the one even prime, 2. To make 59's sum match the others' parity, the 2 must hide behind 59: 59 + 2 = 61 (odd). Then 44 and 38 need odd primes to reach 61 (odd), which is consistent.
Common sum = 61. Behind 44: 61 − 44 = 17 (prime ✓). Behind 38: 61 − 38 = 23 (prime ✓). All six numbers (44, 59, 38, 2, 17, 23) are different, as required.
Average of the hidden primes: (2 + 17 + 23) ÷ 3 = 42 ÷ 3 = 14.
The transferable move: "2 is the only even prime" is a parity wildcard — whenever primes must satisfy an even/odd condition, the 2 is almost always the special case that makes it work. Spot the odd-one-out, and parity tells you where it goes.
