# 2004 AIME II Problems/Problem 3

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

A solid rectangular block is formed by gluing together $N$ congruent 1-cm cubes face to face. When the block is viewed so that three of its faces are visible, exactly $231$ of the 1-cm cubes cannot be seen. Find the smallest possible value of $N.$

## Solution

The $231$ cubes which are not visible must lie below exactly one layer of cubes. Thus, they form a rectangular solid which is one unit shorter in each dimension. If the original block has dimensions $l \times m \times n$, we must have $(l - 1)\times(m-1) \times(n - 1) = 231$. The prime factorization of $231 = 3\cdot7\cdot11$, so we have a variety of possibilities; for instance, $l - 1 = 1$ and $m - 1 = 11$ and $n - 1 = 3 \cdot 7$, among others. However, it should be fairly clear that the way to minimize $l\cdot m\cdot n$ is to make $l$ and $m$ and $n$ as close together as possible, which occurs when the smaller block is $3 \times 7 \times 11$. Then the extra layer makes the entire block $4\times8\times12$, and $N= \boxed{384}$.

An alternate way to visualize the problem is to count the blocks that can be seen and subtract the blocks that cannot be seen. In the given block with dimensions $l\times m \times n$, the three faces have $lm$, $mn$, and $ln$ blocks each. However, $l$ blocks along the first edge, $m$ blocks along the second edge, and $n$ blocks along the third edge were counted twice, so they must be subtracted. After subtracting these three edges, 1 block has not been counted - it was added three times on each face, but subtracted three times on each side. Thus, the total number of visible cubes is $lm+mn+ln-l-m-n+1$, and the total number of invisible cubes is $lmn-lm-mn-ln+l+m+n-1$, which can be factored into $(l-1)(m-1)(n-1)$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 