Can you pair each shaded piece with an unshaded piece of the same size?
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Look for symmetry. For every shaded triangle of the star, is there a matching white triangle the same size in the grid?
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The whole pattern sits on a 4 × 4 grid, so its total area is 16 unit squares.
The star is built from triangles, and each shaded triangle has a congruent white triangle as its partner — the shaded and white regions match up exactly.
So the star covers exactly half of the grid: 8 of the 16 squares, which is 50%.
Buffalo Shuffle-o is a card game in which all the cards are distributed evenly among all players at the start of the game. When Annika and 3 of her friends play Buffalo Shuffle-o, each player is dealt 15 cards. Suppose 2 more friends join the next game. How many cards will be dealt to each player?
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The total number of cards doesn't change. Find that total first.
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First find the total number of cards (it doesn't change), then share it among the new, larger group.
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Annika + 3 friends = 4 players, each dealt 15, so there are 4 × 15 = 60 cards.
With 2 more friends, there are now 4 + 2 = 6 players.
On a street grid you can't cut corners. Each leg's length is just sideways blocks + up-and-down blocks.
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Betty drives along streets, so each leg's length is its sideways blocks plus its up-and-down blocks (you can't cut diagonally). Add the four legs F→A→B→C→F.
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On a street grid, each leg's length = (horizontal blocks) + (vertical blocks). You can't cut diagonals, and as long as you don't backtrack, the route within a leg doesn't matter — only the start and end do.
Read the four legs off the map: F→A = 1 + 2 = 3, A→B = 7 + 3 = 10, B→C = 2 + 4 = 6, C→F = 4 + 1 = 5.
Total: 3 + 10 + 6 + 5 = 24 blocks.
C
24 blocks.
Another way: C is already on the way back (MAA)
Notice C lies on a shortest path from B back to F, so visiting C costs nothing extra. The problem reduces to F → A → B → F.
Sekou writes down the numbers 15, 16, 17, 18, 19. After he erases one of his numbers, the sum of the remaining four numbers is a multiple of 4. Which number did he erase?
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Adding the five up and testing each removal is slow. What does each number have in common with 4?
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Look at each number's remainder when divided by 4. The remainder of the whole sum tells you which one to erase.
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Remainders mod 4: 15→3, 16→0, 17→1, 18→2, 19→3. Their sum is 9, which leaves remainder 1 mod 4.
To make the remaining four sum divisible by 4, erase the one whose remainder is 1 — that's 17.
When two shapes overlap, adding their areas counts the overlap twice. What's the shape of that overlap?
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Inclusion–exclusion: (area of one) + (area of the other) − (area of overlap). The overlap is a square of side 2.5.
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Each rectangle has area 5 × 3 = 15.
Rotated 90° about the midpoint of DC, the second rectangle's lower-left quarter overlaps the first rectangle's lower-right quarter — a 2.5 by 2.5 square (half of DC = 2.5), area 2.52 = 6.25.
The biggest circle that fits is limited by whichever inward corners of the region poke closest to the center.
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Each inward corner sits 1 unit horizontally and 2 units vertically from the center. Use the Pythagorean theorem to get the radius.
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By symmetry, the largest inscribed circle is centered at the region's center. Its radius is the distance from there to the nearest inward-poking corner.
Each such corner is 1 unit across and 2 units up (or down) from the center: distance = √(12 + 22) = √5.
Five distinct integers from 1 to 10 are chosen, and five distinct integers from 11 to 20 are chosen. No two numbers differ by exactly 10. What is the sum of the ten chosen numbers?
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Each chosen low number blocks exactly one high number. How many highs are left to choose from?
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Five chosen lows block five highs — leaving exactly five unblocked highs, which are forced to be picked. They're the highs matching the unchosen lows + 10.
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Each chosen low (say x) blocks the high x + 10. With 5 lows chosen, 5 highs are blocked — so the 5 chosen highs are exactly the 5 unblocked ones: those that are 10 more than the 5 unchosen lows.
Sum of chosen highs = (sum of unchosen lows) + 5 × 10. So (chosen lows) + (chosen highs) = (chosen lows) + (unchosen lows) + 50 = (sum of 1 to 10) + 50.
1 + 2 + … + 10 = 55, so the total is 55 + 50 = 105.
Once both cars are in the middle section, they're going the same speed — the asymmetry is in how they got there.
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Car A reaches the middle in 5/25 = 1/5 hr; car B in 5/20 = 1/4 hr. A has a 1/20-hr head start in the middle section — that's 2 miles.
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Car A's left section is 5 miles at 25 mph → reaches the middle in 5/25 = 1/5 hr. Car B's right section is 5 miles at 20 mph → reaches the middle in 5/20 = 1/4 hr. A enters the middle 1/20 hr before B.
In that 1/20 hr, A travels 40 × 1/20 = 2 miles, so when B enters the middle, A is at mile 7, B at mile 10 — a 3-mile gap.
Both now go 40 mph, closing at 80 mph. They split the 3-mile gap equally: each covers 1.5 miles. A is at 7 + 1.5 = 8.5 miles from A.
Sarika, Dev, and Rajiv are sharing a large block of cheese. They take turns cutting off half of what remains and eating it: first Sarika eats half of the cheese, then Dev eats half of the remaining half, then Rajiv eats half of what remains, then back to Sarika, and so on. They stop when the cheese is too small to see. About what fraction of the original block of cheese does Sarika eat in total?
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Sarika eats on turns 1, 4, 7, … — every third turn. What fraction does each Sarika-bite eat?
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Sarika's bites are 1/2, then 1/16, then 1/128, … — a geometric series with ratio 1/8.
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Each turn eats half of what's left, so the cheese remaining after turn n is 1/2n. Sarika eats at turns 1, 4, 7, …, taking 12, 116, 1128, … of the original block.
Geometric series with first term a = 12 and ratio r = 18.
Look at the largest group of pods that are all directly connected to each other — their grades are forced into a small set.
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Pods A, B, C, F form a 4-clique. The only 4-element subset of {1,…,7} with every pair differing by at least 2 is {1, 3, 5, 7}.
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Among A, B, C, F every pair is directly connected, so all four pairs differ by ≥ 2. The only way to pick 4 numbers from 1–7 with that property is the set {1, 3, 5, 7}.
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.
Imagine adding a phantom coat after the last one. Now the row is a clean repeating pattern.
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After the phantom, you have 36 hooks split into d identical blocks of b hooks each, with b ≥ 2 and d ≥ 2. How many (b, d) factorizations of 36 are there?
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Add a phantom coat after the last real coat. Now 36 hooks form a perfectly repeating pattern: each block is (some empty hooks) + (one coat), of length b ≥ 2.
If there are d blocks, then bd = 36, and the real coat count = d − 1 (we added one).
Both b ≥ 2 (each block has ≥ 1 empty + 1 coat) and d ≥ 2 (at least 1 original coat + the phantom).
36 = 22 × 32 has (2+1)(2+1) = 9 divisors. Removing the two ordered factorizations with a 1 (1×36 and 36×1) leaves 9 − 2 = 7.
By left/right symmetry, the total of right-side areas equals the total of left-side areas. What does each path's left + right equal?
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Per path: area_left + area_right = 25 (the whole diamond). And the number of paths = ways to interleave 5 NE moves with 5 NW moves.
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Let X = the sum of right-side areas. By mirroring every path L ↔ R, the sum of left-side areas across all paths is also X. So 2X = total of (left + right) over all paths.
For each individual path, (left area) + (right area) = 25 (the full 5×5 diamond).
Number of paths: 5 NE moves and 5 NW moves interleaved = 10!5! · 5! = 252.
