Difference between revisions of "2004 AIME II Problems/Problem 3"
I_like_pie (talk | contribs) |
|||
| Line 3: | Line 3: | ||
== Solution == | == 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 <math>3\cdot7\cdot11</math>, 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 <math>4\cdot8\cdot12</math>. Multiplying gives us the answer, <math>N=384</math>. | ||
== See also == | == See also == | ||
* [[2004 AIME II Problems]] | * [[2004 AIME II Problems]] | ||
Revision as of 01:32, 6 November 2006
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 231 of the 1-cm cubes cannot be seen. Find the smallest possible value of 
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 
, 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 
. Multiplying gives us the answer, 
.
