Difference between revisions of "2004 AIME II Problems/Problem 3"

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== See also ==
 
== See also ==
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Revision as of 21:51, 13 February 2007

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

231 cubes cannot be visible, so at least one extra layer of cubes must be on top of these. The prime factorization of 231 is $3\cdot7\cdot11$, and that is the only combination of lengths of sides we have for the smaller block (without the extra layer). The extra layer makes the entire block $4\cdot8\cdot12$. Multiplying gives us the answer, $N=384$.

See also

2004 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions