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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 block is formed by gluing together <math> N </math> congruent 1-cm [[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 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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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>.
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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>.
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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 ==
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{{AIME box|year=2004|num-b=1|num-a=3|n=II}}
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{{AIME box|year=2004|num-b=2|num-a=4|n=II}}
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[[Category:Intermediate Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 11:53, 10 March 2014

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 $N= \boxed{384}$.

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 $l\times m \times n$, the three faces have $lm$, $mn$, and $ln$ blocks each. However, $l$ blocks along the first edge, $m$ blocks along the second edge, and $n$ 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 $lm+mn+ln-l-m-n+1$, and the total number of invisible cubes is $lmn-lm-mn-ln+l+m+n-1$, which can be factored into $(l-1)(m-1)(n-1)$.

See also

2004 AIME II (Problems • Answer Key • Resources)
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Problem 4
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