2004 AIME II Problems/Problem 3
Problem
A solid rectangular block is formed by gluing together congruent 1-cm cubes face to face. When the block is viewed so that three of its faces are visible, exactly of the 1-cm cubes cannot be seen. Find the smallest possible value of
Solution
The 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 , we must have . The prime factorization of , so we have a variety of possibilities; for instance, and and , among others. However, it should be fairly clear that the way to minimize is to make and and as close together as possible, which occurs when the smaller block is . Then the extra layer makes the entire block , and .
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 , the three faces have , , and blocks each. However, blocks along the first edge, blocks along the second edge, and 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 , and the total number of invisible cubes is , which can be factored into .
See also
2004 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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