2004 AIME II Problems/Problem 3

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 $N= \boxed{384}$ .

2004 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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2004 AIME II Problems/Problem 3

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