Difference between revisions of "1977 AHSME Problems/Problem 25"
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We first observe that since there will be more 2s than 5s in <math>1005!</math>, we are looking for the largest <math>n</math> such that <math>5^n</math> divides <math>1005!</math>. We will use the fact that: | We first observe that since there will be more 2s than 5s in <math>1005!</math>, we are looking for the largest <math>n</math> such that <math>5^n</math> divides <math>1005!</math>. We will use the fact that: | ||
Latest revision as of 03:14, 27 November 2021
Problem 25
Determine the largest positive integer 
such that 
is divisible by 
.
Solution
We first observe that since there will be more 2s than 5s in , we are looking for the largest such that 
divides . We will use the fact that:
![\[n = \left \lfloor {\frac{1005}{5^1}}\right \rfloor + \left \lfloor {\frac{1005}{5^2}}\right \rfloor + \left \lfloor {\frac{1005}{5^3}}\right \rfloor \cdots\]](http://latex.artofproblemsolving.com/1/a/2/1a2a9ae5f0cac44d75db08321a83d87e4f9c2fb3.png)
(This is an application of Legendre's formula).
From 
and onwards, 
. Thus, our calculation becomes
![\[n = \left \lfloor {\frac{1005}{5^1}}\right \rfloor + \left \lfloor {\frac{1005}{5^2}}\right \rfloor + \left \lfloor {\frac{1005}{5^3}}\right \rfloor + \left \lfloor {\frac{1005}{5^4}}\right \rfloor\]](http://latex.artofproblemsolving.com/f/f/8/ff8b8686c28898646e7044ccd6e98f295c65ad65.png)
![\[n = \left \lfloor {\frac{1005}{5}}\right \rfloor + \left \lfloor {\frac{1005}{25}}\right \rfloor + \left \lfloor {\frac{1005}{125}}\right \rfloor + \left \lfloor {\frac{1005}{625}}\right \rfloor\]](http://latex.artofproblemsolving.com/2/3/2/232c9785272533dbe8b5b6fc35c26afafdd3d43c.png)
![\[n = 201 + 40 + 8 + 1 = \boxed{250}\]](http://latex.artofproblemsolving.com/f/d/7/fd7bacd3c39d6174a9258ae0b4c9e44aee53e2d2.png)
Since none of the answer choices equal 250, the answer is 
.
- mako17
