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

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== Problem ==
 
== Problem ==
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Let <math> \displaystyle P </math> be the product of the first <math>\displaystyle 100</math> positive odd integers. Find the largest integer <math>\displaystyle k </math> such that <math>\displaystyle P </math> is divisible by <math>\displaystyle 3^k .</math>
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Let <math> \displaystyle P </math> be the product of the first <math>\displaystyle 100</math> [[positive integer | positive]] [[odd integer]]s. Find the largest integer <math>\displaystyle k </math> such that <math>\displaystyle P </math> is divisible by <math>\displaystyle 3^k .</math>
  
 
== Solution ==
 
== Solution ==

Revision as of 16:01, 12 October 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

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

threes in $\displaystyle 200!$ and

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

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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