2024 AMC 10B Problem 25

Problem 25 of 25HarderGeometry Number Theory

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?

Show solution

Solution

Relabel the brick dimensions so the new block has sides

7a,2b,2c.

These must be the three old side lengths

3a,3b,3c,

each increased by

1.

A new side cannot match the old side with the same letter:

7a=3a+1

has no positive integer solution, while

2b=3b+1

and

2c=3c+1

would give negative lengths. Therefore the matching must be one of the two three-cycles. In one orientation,

\begin{aligned} 7a&=3c+1,\\ 2b&=3a+1,\\ 2c&=3b+1. \end{aligned}

The last two equations give

b=\frac{3a+1}{2}

and

c=\frac{9a+5}{4}.

Substituting into the first gives

28a=27a+19,

(a,b,c)=(19,29,44).

The other cycle merely exchanges

b

and

c.

These lengths are pairwise relatively prime, and

a+b+c=19+29+44=92.

Thus, E is the correct answer.