2024 AMC 10B Problem 25

The old block has dimensions $3a, 3b, 3c$. In the new block the bricks are all oriented the same way, so its dimensions are $7p, 2q, 2r$ where $(p, q, r)$ is some ordering of $(a, b, c)$. Each new dimension is exactly $1$ more than one old dimension, matching them up one to one.

A new dimension cannot be matched with the old dimension built from the same edge: $2p = 3p + 1$ and $7p = 3p + 1$ have no positive solutions. So the matching is a derangement of three items, which must be a $3$-cycle. Naming the edges so that the $7$-direction uses $p$, the equations are
$$7p = 3q + 1, \qquad 2q = 3r + 1, \qquad 2r = 3p + 1$$
(the other $3$-cycle is the same system with $q$ and $r$ swapped).

Substitute upward: $r = \frac{3p + 1}{2}$, then $q = \frac{3r + 1}{2} = \frac{9p + 5}{4}$, then
$$7p = 3 \cdot \frac{9p + 5}{4} + 1 = \frac{27p + 19}{4} \quad\Longrightarrow\quad 28p = 27p + 19 \quad\Longrightarrow\quad p = 19.$$
Then $r = 29$ and $q = 44$. These are pairwise coprime, as required, and
$$a + b + c = 19 + 29 + 44 = 92.$$

The answer is $\boxed{\textbf{(E)}\ 92}$.

The physical setup hides a linear system, but the real insight is deciding which new dimension corresponds to which old one. Ruling out "same edge" pairings by a one-line impossibility argument leaves a single cyclic structure, and the sum $a + b + c$ is the same regardless of how the letters are attached. Read figures for structure ($7$ bricks along one direction), not for orientation promises.

Add the three equations: $7p + 2q + 2r = 3(p + q + r) + 3$, so $q + r = 4p - 3$ and $a + b + c = 5p - 3 \equiv 2 \pmod 5$. Among the choices only $92$ leaves remainder $2$ when divided by $5$.

• Assuming the bricks keep their orientation, so that the new width 2a should be 3a + 1, which is impossible and stalls the whole problem.
• Solving the system with a fraction error (e.g. $2c = 3b + 1$ handled as $c = 3b + 1$) and landing on a non-integer, then guessing.