Question

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 $a + b + c?$

$88$

$89$

$90$

$91$

$92$

Solution:

The $3 \times 3 \times 3$ block has sides $3a, 3b, 3c.$ The $2 \times 2 \times 7$ block has sides $2u, 2v, 7w,$ where $(u, v, w)$ is some permutation of $(a, b, c).$ Each new side beats its old counterpart by $1,$ so the multiset $\{3a + 1, 3b + 1, 3c + 1\}$ equals $\{2u, 2v, 7w\}.$ Try matching the $7$ -side to $3(\text{one value}) + 1.$ With $\{a, b, c\} = \{19, 29, 44\}$ it all lines up: $7 \cdot 19 = 133 = 3 \cdot 44 + 1,$ $2 \cdot 29 = 58 = 3 \cdot 19 + 1,$ and $2 \cdot 44 = 88 = 3 \cdot 29 + 1.$ Those are pairwise coprime, so $a + b + c = 19 + 29 + 44 = 92.$ Thus, E is the correct answer.

2024 AMC 10B Problem 25

Solution:

Problem 25 in Other Years