Difference between revisions of "1977 AHSME Problems/Problem 25"
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== Problem 25 == | == Problem 25 == | ||
Determine the largest positive integer <math>n</math> such that <math>1005!</math> is divisible by <math>10^n</math>. | Determine the largest positive integer <math>n</math> such that <math>1005!</math> is divisible by <math>10^n</math>. | ||
| + | |||
<math> | <math> | ||
\textbf{(A) }102\qquad | \textbf{(A) }102\qquad | ||
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\textbf{(C) }249\qquad | \textbf{(C) }249\qquad | ||
\textbf{(D) }502\qquad | \textbf{(D) }502\qquad | ||
| + | \textbf{(E) }\text{none of the above}\qquad | ||
</math> | </math> | ||
| − | \\\\ | + | |
| + | |||
| + | === Solution === | ||
| + | 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: | ||
| + | |||
| + | <cmath>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</cmath> | ||
| + | |||
| + | (This is an application of Legendre's formula). | ||
| + | |||
| + | From <math>k=5</math> and onwards, <math>\left \lfloor {\frac{1005}{5^k}}\right \rfloor = 0</math>. Thus, our calculation becomes | ||
| + | |||
| + | <cmath>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</cmath> | ||
| + | |||
| + | <cmath>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</cmath> | ||
| + | |||
| + | <cmath>n = 201 + 40 + 8 + 1 = \boxed{250}</cmath> | ||
| + | |||
| + | Since none of the answer choices equal 250, the answer is <math>\boxed{\textbf{(E)}}</math>. | ||
| + | |||
| + | - mako17 | ||
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
