Difference between revisions of "1977 AHSME Problems/Problem 25"

(Problem 25)
(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>.
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<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) }none of these\qquad
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\textbf{(E) }\text{none of the above}\qquad
 
</math>
 
</math>
\\\\
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=== Solution ===
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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:
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<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>
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(This is an application of Legendre's formula).
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From <math>k=5</math> and onwards, <math>\left \lfloor {\frac{1005}{5^k}}\right \rfloor = 0</math>. Thus, our calculation becomes
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<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>
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<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>
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<cmath>n = 201 + 40 + 8 + 1 = \boxed{250}</cmath>
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Since none of the answer choices equal 250, the answer is <math>\boxed{\textbf{(E)}}</math>.
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- mako17

Latest revision as of 04:14, 27 November 2021

Problem 25

Determine the largest positive integer $n$ such that $1005!$ is divisible by $10^n$.

$\textbf{(A) }102\qquad \textbf{(B) }112\qquad \textbf{(C) }249\qquad \textbf{(D) }502\qquad \textbf{(E) }\text{none of the above}\qquad$


Solution

We first observe that since there will be more 2s than 5s in $1005!$, we are looking for the largest $n$ such that $5^n$ divides $1005!$. We will use the fact that:

\[n = \left \lfloor {\frac{1005}{5^1}}\right \rfloor + \left \lfloor {\frac{1005}{5^2}}\right \rfloor + \left \lfloor {\frac{1005}{5^3}}\right \rfloor \cdots\]

(This is an application of Legendre's formula).

From $k=5$ and onwards, $\left \lfloor {\frac{1005}{5^k}}\right \rfloor = 0$. Thus, our calculation becomes

\[n = \left \lfloor {\frac{1005}{5^1}}\right \rfloor + \left \lfloor {\frac{1005}{5^2}}\right \rfloor + \left \lfloor {\frac{1005}{5^3}}\right \rfloor + \left \lfloor {\frac{1005}{5^4}}\right \rfloor\]

\[n = \left \lfloor {\frac{1005}{5}}\right \rfloor + \left \lfloor {\frac{1005}{25}}\right \rfloor + \left \lfloor {\frac{1005}{125}}\right \rfloor + \left \lfloor {\frac{1005}{625}}\right \rfloor\]

\[n = 201 + 40 + 8 + 1 = \boxed{250}\]

Since none of the answer choices equal 250, the answer is $\boxed{\textbf{(E)}}$.

- mako17

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