Difference between revisions of "2004 AIME II Problems/Problem 3"
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== Problem == | == Problem == | ||
− | A solid [[rectangular solid | rectangular block]] is formed by gluing together <math> N </math> [[congruent (geometry) | congruent]] 1-cm [[cube (geometry) | cube]]s [[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 <math> N. </math> | + | A solid [[rectangular solid | rectangular block]] is formed by gluing together <math> N </math> [[congruent (geometry) | congruent]] 1-cm [[cube (geometry) | cube]]s [[face]] to face. When the block is viewed so that three of its faces are visible, exactly <math>231</math> of the 1-cm cubes cannot be seen. Find the smallest possible value of <math> N. </math> |
== Solution == | == 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 <math>l \times m \times n</math>, we must have <math>(l - 1)\times(m-1) \times(n - 1) = 231</math>. The [[prime factorization]] of | + | The <math>231</math> 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 <math>l \times m \times n</math>, we must have <math>(l - 1)\times(m-1) \times(n - 1) = 231</math>. The [[prime factorization]] of <math>231 = 3\cdot7\cdot11</math>, so we have a variety of possibilities; for instance, <math>l - 1 = 1</math> and <math>m - 1 = 11</math> and <math>n - 1 = 3 \cdot 7</math>, among others. However, it should be fairly clear that the way to minimize <math>l\cdot m\cdot n</math> is to make <math>l</math> and <math>m</math> and <math>n</math> as close together as possible, which occurs when the smaller block is <math>3 \times 7 \times 11</math>. Then the extra layer makes the entire block <math>4\times8\times12</math>, and <math>N= \boxed{384}</math>. |
== See also == | == See also == | ||
{{AIME box|year=2004|num-b=2|num-a=4|n=II}} | {{AIME box|year=2004|num-b=2|num-a=4|n=II}} | ||
+ | |||
+ | [[Category:Intermediate Number Theory Problems]] |
Revision as of 19:47, 25 July 2008
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 .
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 | ||
All AIME Problems and Solutions |