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

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== Problem ==
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#REDIRECT [[2006 AIME I Problems/Problem 3]]
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>
 
 
 
== Solution ==
 
 
 
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>
 
 
 
Hence, we seek the number of threes in <math>\displaystyle 200!</math> decreased by the number of threes in <math>\displaystyle 100!.</math>
 
 
 
There are
 
 
 
<math>\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</math>
 
threes in <math>\displaystyle 200!</math> and
 
<math>\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 </math>
 
 
 
threes in <math>\displaystyle 100!</math>
 
 
 
Therefore, we have a total of <math>\displaystyle 97-48=049</math> threes.
 
 
 
 
 
== See also ==
 
*[[2006 AIME II Problems]]
 
 
 
[[Category:Intermediate Number Theory Problems]]
 

Latest revision as of 11:21, 14 June 2009