AMC 10 · 2024 · #25

Grade 8 number-theory

Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Integrality / Divisibility Screen

gcdsystems-of-equationsprime-factorizationspatial-visualizationvolume-rectangular-prism identify-subproblemsconvert-to-algebracaseworkguess-and-check ↑ Prerequisites: gcdsystems-of-equationsprime-factorizationvolume-rectangular-prism

📏 Long solution 💡 4 insights 📊 Diagram

Problem

27 identical a × b × c bricks (with a, b, c pairwise coprime positive integers) stack into a 3 × 3 × 3 block. Adding one more brick gives 28 bricks that re-stack into a 2 × 2 × 7 block whose three dimensions are each exactly 1 unit larger than the corresponding dimensions of the first block. Find a + b + c.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Tool #1 (Draw a Diagram) starts: a 3 × 3 × 3 stack of bricks has outer dimensions 3a, 3b, 3c in some axis order, and a 2 × 2 × 7 stack has 2x, 2y, 7z in some axis order — that geometry is one quick sketch. Tool #7 (Identify Subproblems) splits the work into three independent matchings (one per axis), each of the form "new outer dim = old outer dim + 1", which Tool #13 (Convert to Algebra) turns into three linear equations. Tool #14 (Use Cases) handles the ambiguity in which brick edge plays the role of "7z" — for the 7 to give an integer answer, 7z = 3 · (some letter) + 1 must be solvable, and only one cyclic assignment (3a+1 = 2b, 3b+1 = 2c, 3c+1 = 7a) leads to small positive integers. Tool #6 (Guess and Check) verifies the answer at the end against the pairwise-coprime requirement.

1STEP 1

The 3 × 3 × 3 stack has outer dims 3a, 3b, 3c; the 2 × 2 × 7 stack has 2x, 2y, 7z with {x, y, z} a permutation of {a, b, c}.

old: {3a, 3b, 3c}, new: {2x, 2y, 7z}, {x,y,z} = {a,b,c}

2STEP 2

Each new dim is an old dim +1, so {3a+1, 3b+1, 3c+1} = {2x, 2y, 7z} as multisets; the lone 7z slot pins the match.

{3a + 1, 3b + 1, 3c + 1} = {2x, 2y, 7z}, {x,y,z} ∼ {a,b,c}

3STEP 3

Only the cycle 3a+1 = 2b, 3b+1 = 2c, 3c+1 = 7a gives positive integers; other matches repeat a letter or force 6a = 1.

3a + 1 = 2b ; 3b + 1 = 2c ; 3c + 1 = 7a

4STEP 4

Substitute down: b = $\frac{3a+1}{2}$ , c = $\frac{9a+5}{4}$ , then 3c+1 = 7a becomes 27a + 19 = 28a, giving a = 19.

b =

\frac{3a+1}{2}

, c =

\frac{9a+5}{4}

\frac{27a + 19}{4}

= 7a → a = 19

5STEP 5

Back-substitute: b = $\frac{58}{2}$ = 29, c = $\frac{176}{4}$ = 44, so the brick is 19 × 29 × 44.

a = 19, b = 29, c = 44

6STEP 6

Check: 19, 29, 44 are pairwise coprime, and {57, 87, 132} + 1 = {58, 88, 133} matches {2b, 2c, 7a}.

{57, 87, 132} + 1 = {58, 88, 133} = {2b, 2c, 7a} ✓

7STEP 7

Add them: a + b + c = 19 + 29 + 44 = 92, choice (E).

a + b + c = 19 + 29 + 44 = 92 → (E)

Answer

The numbers pass three independent checks. (1) The brick 19 × 29 × 44 is plausible for AMC-final-problem scale — neither tiny (which would make the picture trivial) nor huge (which would make arithmetic impossible). (2) Pairwise coprimality holds, exactly because the problem demanded it. (3) Both blocks' outer dimensions match: {57, 87, 132} vs {58, 88, 133}, with every entry exactly 1 bigger. The answer 92 is the top choice (E); the nearby distractors 88, 89, 90, 91 correspond to small arithmetic slips (19 + 29 + 40, etc.) that this verification rules out.

💡Key takeaway

Two brick stacks differing by one along each axis become three small equations once you label dimensions: 3a+1=2b, 3b+1=2c, 3c+1=7a. Solve in cascade — a = 19, b = 29, c = 44 — and the sum 92 closes (E).

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