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Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 2"
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2. Let <math>\star (x)</math> be the sum of the digits of a positive integer <math>x</math>. <math>\mathcal{S}</math> is the set of positive integers such that for all elements <math>n</math> in <math>\mathcal{S}</math>, we have that <math>\star (n)=12</math> and <math>0\le n< 10^{7}</math>. If <math>m</math> is the number of elements in <math>\mathcal{S}</math>, compute <math>\star(m)</math>.
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Let <math>\star (x)</math> be the sum of the digits of a positive integer <math>x</math>. <math>\mathcal{S}</math> is the set of positive integers such that for all elements <math>n</math> in <math>\mathcal{S}</math>, we have that <math>\star (n)=12</math> and <math>0\le n< 10^{7}</math>. If <math>m</math> is the number of elements in <math>\mathcal{S}</math>, compute <math>\star(m)</math>.
  
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[[Mock AIME 1 2006-2007]]
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==Solution==
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Equivalently, we need to place 12 indistinguishable balls into 7 distinguishable boxes so that no box contains more than 9 balls.  There are <math>{12 + 7 - 1 \choose 7 - 1} = {18 \choose 6} = 18,564</math> ways to place 12 objects into 7 boxes.  Of these, 7 place all 12 into a single box.  <math>7 \cdot 6 = 42</math> place 11 in one box and 1 in a second.  <math>7 \cdot 6 = 42</math> place 10 into one box and 2 into a second.  <math>7 \cdot \frac{6\cdot 5}{2} = 105</math> place 10 into one box and 1 into each of two others.  Thus, this gives us <math>m = 18564 - 7 - 42 - 42 - 105 = 18368</math> so <math>\star(m) = 1 + 8 + 3 + 6 + 8 = 026</math>.
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*[[Mock AIME 1 2006-2007 Problems/Problem 1 | Previous Problem]]
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*[[Mock AIME 1 2006-2007 Problems/Problem 3 | Next Problem]]
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*[[Mock AIME 1 2006-2007]]
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[[Category:Intermediate Combinatorics Problems]]

Latest revision as of 15:53, 3 April 2012

Let $\star (x)$ be the sum of the digits of a positive integer $x$. $\mathcal{S}$ is the set of positive integers such that for all elements $n$ in $\mathcal{S}$, we have that $\star (n)=12$ and $0\le n< 10^{7}$. If $m$ is the number of elements in $\mathcal{S}$, compute $\star(m)$.

Solution

Equivalently, we need to place 12 indistinguishable balls into 7 distinguishable boxes so that no box contains more than 9 balls. There are ${12 + 7 - 1 \choose 7 - 1} = {18 \choose 6} = 18,564$ ways to place 12 objects into 7 boxes. Of these, 7 place all 12 into a single box. $7 \cdot 6 = 42$ place 11 in one box and 1 in a second. $7 \cdot 6 = 42$ place 10 into one box and 2 into a second. $7 \cdot \frac{6\cdot 5}{2} = 105$ place 10 into one box and 1 into each of two others. Thus, this gives us $m = 18564 - 7 - 42 - 42 - 105 = 18368$ so $\star(m) = 1 + 8 + 3 + 6 + 8 = 026$.

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