2024 AMC 10B Problem 25

Step 1: write the sizes.

1. The old block is $3$ bricks in each direction, so its sides are $3a$, $3b$, $3c$.
2. In the new block every brick points the same way. Say the direction with $7$ bricks runs along the brick edge $a$, and the two directions with $2$ bricks run along $b$ and $c$. Then the new sides are $7a$, $2b$, $2c$.
3. Each new side is $1$ more than one of the old sides, and each old side is used once.

Step 2: match new sides to old sides.

1. A new side cannot match the old side made from the same brick edge. For example $7a = 3a + 1$ would give $a = \frac14$, and $2b = 3b + 1$ would give $b = -1$. Neither is a positive whole number.
2. So $7a$ must match $3b$ or $3c$. The names $b$ and $c$ are interchangeable so far, so say $7a = 3c + 1$.
3. Now $2b$ must match $3a$ or $3b$. It cannot be $3b$ (step 4), so $2b = 3a + 1$.
4. That leaves $2c = 3b + 1$.

Step 3: solve.

8. The three equations are
$$7a = 3c + 1, \qquad 2b = 3a + 1, \qquad 2c = 3b + 1.$$
9. From the second, $b = \frac{3a + 1}{2}$. Put that in the third: $2c = \frac{9a + 3}{2} + 1 = \frac{9a + 5}{2}$, so $c = \frac{9a + 5}{4}$.
10. Put this in the first: $7a = \frac{27a + 15}{4} + 1 = \frac{27a + 19}{4}$. Multiply by $4$: $28a = 27a + 19$, so $a = 19$.
11. Then $b = \frac{57 + 1}{2} = 29$ and $c = \frac{87 + 1}{2} = 44$. Check the first equation: $7 \cdot 19 = 133 = 3 \cdot 44 + 1$. Also $19$, $29$, $44$ share no common factor in any pair, as the problem requires.
12. So $a + b + c = 19 + 29 + 44 = 92$.

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

The real puzzle is deciding which new side grows from which old side. One quick impossibility check (a side cannot grow from its own brick edge) leaves only one way to match them, and after that it is three simple equations. Picture the blocks by their structure (how many bricks run in each direction) rather than assuming the bricks keep their old positions.

Add the three equations: $7a + 2b + 2c = 3a + 3b + 3c + 3$. This simplifies to $b + c = 4a - 3$, so $a + b + c = 5a - 3$. That is $2$ more than a multiple of $5$ (since $5a - 3 = 5(a - 1) + 2$). Among the choices, only $92$ is $2$ more than a multiple of $5$.

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 $\boxed{\textbf{(E)}\ 92}$.