2006 AIME A Problems/Problem 3: Difference between revisions
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== Problem == | == Problem == | ||
Let <math> P </math> be the product of the first 100 positive odd integers. Find the largest integer <math> k </math> such that <math> P </math> is divisible by <math> 3^k </math> | 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 == | == 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 <center><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></center> | |||
threes in <math>\displaystyle 200!</math> and | |||
<center><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></center> | |||
threes in <math>\displaystyle 100!</math> | |||
Therefore, we have a total of <math>\displaystyle 97-48=049</math> threes. | |||
== See also == | == See also == | ||
Revision as of 13:28, 24 July 2006
Problem
Let 
be the product of the first 
positive odd integers. Find the largest integer 
such that is divisible by 
Solution
Note that the product of the first positive odd integers can be written as 
Hence, we seek the number of threes in 
decreased by the number of threes in 
There are
threes in and
threes in 
Therefore, we have a total of 
threes.
