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

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== Problem ==
 
== Problem ==
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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>
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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 ==
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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 231 is <math>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=384</math>.
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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}}
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[[Category:Intermediate Number Theory Problems]]

Revision as of 18:47, 25 July 2008

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

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 $l \times m \times n$, we must have $(l - 1)\times(m-1) \times(n - 1) = 231$. The prime factorization of $231 = 3\cdot7\cdot11$, so we have a variety of possibilities; for instance, $l - 1 = 1$ and $m - 1 = 11$ and $n - 1 = 3 \cdot 7$, among others. However, it should be fairly clear that the way to minimize $l\cdot m\cdot n$ is to make $l$ and $m$ and $n$ as close together as possible, which occurs when the smaller block is $3 \times 7 \times 11$. Then the extra layer makes the entire block $4\times8\times12$, and $N= \boxed{384}$.

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
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