AMC 10 · 2024 · #25

Grade 8 number-theory

Archetype Multiplicative Ratio with Integrality Constraints

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

Each of $27$ bricks (right rectangular prisms) has dimensions $a \times b \times c$ , where $a$ , $b$ , and $c$ are pairwise relatively prime positive integers. These bricks are arranged to form a $3 \times 3 \times 3$ block, as shown on the left below. A $28$ th brick with the same dimensions is introduced, and these bricks are reconfigured into a $2 \times 2 \times 7$ block, shown on the right. The new block is $1$ unit taller, $1$ unit wider, and $1$ unit deeper than the old one. What is $a + b + c$ ?

$\textbf{(A) }88 \qquad \textbf{(B) }89 \qquad \textbf{(C) }90 \qquad \textbf{(D) }91 \qquad \textbf{(E) }92 \qquad$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

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

Givens: Brick dimensions: $a \times b \times c$ with $\gcd(a,b) = \gcd(b,c) = \gcd(a,c) = 1$; $3 \times 3 \times 3$ stacking ($27$ bricks): outer dimensions are some permutation of $\{3a, 3b, 3c\}$; $2 \times 2 \times 7$ stacking ($28$ bricks): outer dimensions are some permutation of $\{2x, 2y, 7z\}$ where $\{x, y, z\} = \{a, b, c\}$; Each new outer dimension is $1$ unit greater than the matching old outer dimension; Answer choices: (A) $88$, (B) $89$, (C) $90$, (D) $91$, (E) $92$

Unknowns: $a + b + c$ — and along the way, the individual values $a, b, c$

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #1 Draw a Diagram, #13 Convert to Algebra, #14 Use Cases, #6 Guess and Check

Tool #1 (Draw a Diagram) starts: a $3 \times 3 \times 3$ stack of bricks has outer dimensions $3a, 3b, 3c$ in some axis order, and a $2 \times 2 \times 7$ 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, $7z = 3 \cdot (\text{some letter}) + 1$ 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.

Execute — Answer: E

#1 Draw a Diagram 5.MD.C.5 Step 1

Read the dimensions off the picture.
In the $3 \times 3 \times 3$ stack, every axis runs $3$ bricks long, so the three outer dimensions are $3a$, $3b$, $3c$ in some axis order.
In the $2 \times 2 \times 7$ stack, two axes run $2$ bricks long and one runs $7$ bricks long, so the three outer dimensions are $2x$, $2y$, $7z$ where $\{x, y, z\}$ is some permutation of $\{a, b, c\}$.

$$\text{old: } \{3a, 3b, 3c\}, \quad \text{new: } \{2x, 2y, 7z\}, \; \{x,y,z\} = \{a,b,c\}$$

💡 Grade 5 "volume by stacking unit boxes" — count bricks per axis to read the outer length.

#13 Convert to Algebra 8.EE.C.8 Step 2

Convert "each new dimension is $1$ unit larger" into three equations.
The set $\{3a+1, 3b+1, 3c+1\}$ (old dims, each $+1$) must equal the set $\{2x, 2y, 7z\}$ (new dims).
The $7z$ slot is the constraining one: it must equal one of $3a+1, 3b+1, 3c+1$ — and the other two slots ($2x, 2y$) consume the remaining two.

$$\{3a + 1, \, 3b + 1, \, 3c + 1\} = \{2x, 2y, 7z\}, \quad \{x,y,z\} \sim \{a,b,c\}$$

💡 Grade 8 system-of-equations setup: a set equality is three equations once we match elements pairwise.

#14 Use Cases 8.EE.C.8 Step 3

Pick the cyclic assignment $3a+1 = 2b, \; 3b+1 = 2c, \; 3c+1 = 7a$.
(Other assignments either repeat a letter — forbidden since $a, b, c$ are distinct — or give equations like $3a+1 = 7a$, i.e.
$6a = 1$, with no positive integer solution.
The cycle $a \to b \to c \to a$ is the only one consistent with positive integers.)

$$\begin{cases} 3a + 1 = 2b \\ 3b + 1 = 2c \\ 3c + 1 = 7a \end{cases}$$

💡 Grade 8 case selection: the $7$-times slot is rare, so trying it first ($7a$ vs $7b$ vs $7c$) quickly narrows to one workable cycle.

#13 Convert to Algebra 8.EE.C.8 Step 4

Solve the system by substitution.
From (1): $b = \dfrac{3a + 1}{2}$.
Plug into (2): $3 \cdot \dfrac{3a + 1}{2} + 1 = 2c$, so $\dfrac{9a + 3 + 2}{2} = 2c$, giving $c = \dfrac{9a + 5}{4}$.
Plug into (3): $3 \cdot \dfrac{9a + 5}{4} + 1 = 7a$, so $\dfrac{27a + 15 + 4}{4} = 7a$, i.e.
$27a + 19 = 28a$, giving $a = 19$.

$$b = \tfrac{3a+1}{2}, \; c = \tfrac{9a+5}{4}, \; \tfrac{27a + 19}{4} = 7a \Rightarrow a = 19$$

💡 Grade 8 substitute and clear denominators — three linear equations cascade to one number.

#13 Convert to Algebra 8.EE.C.7 Step 5

Back-substitute to find $b$ and $c$.
$b = \tfrac{3 \cdot 19 + 1}{2} = \tfrac{58}{2} = 29$.
$c = \tfrac{9 \cdot 19 + 5}{4} = \tfrac{176}{4} = 44$.
So the brick is $19 \times 29 \times 44$.

$$a = 19, \; b = 29, \; c = 44$$

💡 Grade 8 linear-equation arithmetic — one value cascades into the others.

#6 Guess and Check 6.NS.B.4 Step 6

Verify the constraints.
Pairwise coprime: $\gcd(19, 29) = 1$ (both prime), $\gcd(19, 44) = 1$ ($19$ prime, $44 = 2^2 \cdot 11$), $\gcd(29, 44) = 1$ ($29$ prime, $44 = 2^2 \cdot 11$).
Dimension match: old block $\{3 \cdot 19, 3 \cdot 29, 3 \cdot 44\} = \{57, 87, 132\}$; new block $\{7 \cdot 19, 2 \cdot 29, 2 \cdot 44\} = \{133, 58, 88\}$.
Pair them up: $57 + 1 = 58$, $87 + 1 = 88$, $132 + 1 = 133$.
All three increases by $1$ check.

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

💡 Grade 6 GCF check + numerical verification — every original constraint passes.

#7 Identify Subproblems 4.NBT.B.4 Step 7

Compute the requested sum.

$$a + b + c = 19 + 29 + 44 = 92 \;\Rightarrow\; \textbf{(E)}$$

💡 Grade 4 three-number addition closes the problem.

[1] #1 5.MD.C.5 Read the dimensions off the picture. In the $3 \times 3 \times 3$ stack, every a

[2] #13 8.EE.C.8 Convert "each new dimension is $1$ unit larger" into three equations. The set $\

[3] #14 8.EE.C.8 Pick the cyclic assignment $3a+1 = 2b, \; 3b+1 = 2c, \; 3c+1 = 7a$. (Other assig

[4] #13 8.EE.C.8 Solve the system by substitution. From (1): $b = \dfrac{3a + 1}{2}$. Plug into (

[5] #13 8.EE.C.7 Back-substitute to find $b$ and $c$. $b = \tfrac{3 \cdot 19 + 1}{2} = \tfrac{58}

[6] #6 6.NS.B.4 Verify the constraints. Pairwise coprime: $\gcd(19, 29) = 1$ (both prime), $\gcd

[7] #7 4.NBT.B.4 Compute the requested sum.

Review

Reasonableness: The numbers pass three independent checks. (1) The brick $19 \times 29 \times 44$ 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.

Alternative: Tool #6 (Guess and Check) on the answer choices. The sum $a + b + c \in \{88, 89, 90, 91, 92\}$ is small, and the smallest dimension $a$ satisfies $3a + 1 = 2b$ (so $a$ odd) plus $3c + 1 = 7a$ (so $7a \equiv 1 \pmod 3$, i.e. $a \equiv 1 \pmod 3$). The smallest odd $a \equiv 1 \pmod 3$ giving integer $b, c$ via the cascade is $a = 19$: check $b = 29$, $c = 44$, sum $92$. The constraints zero in on (E) without solving the full system algebraically — useful as a sanity verification.

CCSS standards used (min grade 8)

5.MD.C.5 Relate volume to the operations of multiplication and addition and solve real-world and mathematical problems involving volume (Reading the outer dimensions of a $3 \times 3 \times 3$ brick stack as $3a, 3b, 3c$ and a $2 \times 2 \times 7$ stack as $2x, 2y, 7z$ by counting bricks per axis.)
8.EE.C.7 Solve linear equations in one variable (Solving the reduced equation $27a + 19 = 28a$ to obtain $a = 19$, then back-substituting to get $b = 29$ and $c = 44$.)
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations (Setting up and chaining the three simultaneous equations $3a+1 = 2b$, $3b+1 = 2c$, $3c+1 = 7a$ from the set-equality $\{3a+1, 3b+1, 3c+1\} = \{2b, 2c, 7a\}$.)
6.NS.B.4 Find the greatest common factor of two whole numbers less than or equal to 100 (Verifying $\gcd(19,29) = \gcd(19,44) = \gcd(29,44) = 1$ to confirm pairwise coprimality of the recovered brick dimensions.)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers using the standard algorithm (Computing $a + b + c = 19 + 29 + 44 = 92$.)

⭐ 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).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 8)

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