Difference between revisions of "2004 AIME II Problems/Problem 3"
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== Problem == | == Problem == | ||
| − | A solid rectangular block is formed by gluing together <math> N </math> congruent 1-cm | + | A 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 == | ||
| − | 231 cubes | + | 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>. |
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| + | 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 <math>l\times m \times n</math>, the three faces have <math>lm</math>, <math>mn</math>, and <math>ln</math> blocks each. However, <math>l</math> blocks along the first edge, <math>m</math> blocks along the second edge, and <math>n</math> 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 <math>lm+mn+ln-l-m-n+1</math>, and the total number of invisible cubes is <math>lmn-lm-mn-ln+l+m+n-1</math>, which can be factored into <math>(l-1)(m-1)(n-1)</math>. | ||
== See also == | == See also == | ||
| − | + | {{AIME box|year=2004|num-b=2|num-a=4|n=II}} | |
| − | + | ||
| − | + | [[Category:Intermediate Number Theory Problems]] | |
| + | {{MAA Notice}} | ||
Latest revision as of 10:53, 10 March 2014
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 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
