|
|
| (10 intermediate revisions by 5 users not shown) |
| Line 1: |
Line 1: |
| − | == Problem ==
| + | #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]]
| |
