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 | ||
Line 6: | Line 7: | ||
\textbf{(C) }249\qquad | \textbf{(C) }249\qquad | ||
\textbf{(D) }502\qquad | \textbf{(D) }502\qquad | ||
− | \textbf{(E) }none of | + | \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:
(This is an application of Legendre's formula).
From and onwards, . Thus, our calculation becomes
Since none of the answer choices equal 250, the answer is .
- mako17