2024 AMC 10B Problems/Problem 25

Revision as of 00:26, 14 November 2024 by Lprado (talk | contribs)

Solution 1

The $3x3x3$ block has side lengths of $3a, 3b, 3c$. The $2x2x7$ block has side lengths of $2b, 2c, 7a$.

We can create the following system of equations, knowing that the new block has 1 unit taller, deeper, and wider than the original: \[3a+1 = 2b\] \[3b+1=2c\] \[3c+1=7a\]

Adding all the equations together, we get $b+c+3 = 4a$. Adding $a-3$ to both sides, we get $a+b+c = 4a-3$. The question states that $a,b,c$ are all relatively prime positive integers. Therefore, our answer must be congruent to $2 \pmod{5}$. The only answer choice satisfying this is $\boxed{92}$. ~lprado