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Difference between revisions of "2006 AIME A Problems/Problem 3"

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== Solution ==
 
== Solution ==
  
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Note that the product of the first <math>\displaystyle 100</math> positive odd integers can be written as <math>\displaystyle 1\cdot 3\cdot 5\cdot 7\cdots 195\cdot 197\cdot 199=\frac{200!}{2(100)!}.</math>
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Note that the product of the first <math>\displaystyle 100</math> positive odd integers can be written as <math>\displaystyle 1\cdot 3\cdot 5\cdot 7\cdots 195\cdot 197\cdot 199=\frac{1\cdot 2\cdots200}{2\cdot4\cdots200} = \frac{200!}{2^{100}\cdot 100!}</math>
  
 
Hence, we seek the number of threes in <math>\displaystyle 200!</math> decreased by the number of threes in <math>\displaystyle 100!.</math>
 
Hence, we seek the number of threes in <math>\displaystyle 200!</math> decreased by the number of threes in <math>\displaystyle 100!.</math>
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Therefore, we have a total of <math>\displaystyle 97-48=049</math> threes.
 
Therefore, we have a total of <math>\displaystyle 97-48=049</math> threes.
  
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For more information, see also [[Factorial#Prime factorization| prime factorizations of a factorial]].
  
 
== See also ==
 
== See also ==

Revision as of 14:35, 24 July 2006

Problem

Let $\displaystyle P$ be the product of the first $\displaystyle 100$ positive odd integers. Find the largest integer $\displaystyle k$ such that $\displaystyle P$ is divisible by $\displaystyle 3^k .$

Solution

Note that the product of the first $\displaystyle 100$ positive odd integers can be written as $\displaystyle 1\cdot 3\cdot 5\cdot 7\cdots 195\cdot 197\cdot 199=\frac{1\cdot 2\cdots200}{2\cdot4\cdots200} = \frac{200!}{2^{100}\cdot 100!}$

Hence, we seek the number of threes in $\displaystyle 200!$ decreased by the number of threes in $\displaystyle 100!.$

There are

$\displaystyle \lfloor \frac{200}{3}\rfloor+\lfloor\frac{200}{9}\rfloor+\lfloor \frac{200}{27}\rfloor+\lfloor\frac{200}{81}\rfloor =66+22+7+2=97$

threes in $\displaystyle 200!$ and

$\displaystyle \lfloor \frac{100}{3}\rfloor+\lfloor\frac{100}{9}\rfloor+\lfloor \frac{100}{27}\rfloor+\lfloor\frac{100}{81}\rfloor=33+11+3+1=48$

threes in $\displaystyle 100!$

Therefore, we have a total of $\displaystyle 97-48=049$ threes.

For more information, see also prime factorizations of a factorial.

See also

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