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Difference between revisions of "2005 iTest Problems/Problem 37"

(Created page with "Since we want to find the number of zeros at the end of <math>209!</math>, it is the same as finding the largest value of <math>n</math> such that <math>10^n</math> is a divis...")
 
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Since we want to find the number of zeros at the end of <math>209!</math>, it is the same as finding the largest value of <math>n</math> such that <math>10^n</math> is a divisor of <math>209!.</math> Since <math>10=2\cdot5</math> and there are more factors of <math>2</math> than <math>5,</math> finding the number of zeros at the end of <math>209!</math> is the same as finding the largest value of <math>m</math> such that <math>5^m</math> that is a divisor of <math>209!.</math> We then can use the floor function to find the factors of <math>5</math> in <math>209!.</math>
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==Problem==
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How many zeroes appear at the end of <math>209</math> factorial?
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==Solution 1==
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Since we want to find the number of zeros at the end of <math>209!</math>, it is the same as finding the largest value of <math>n</math> such that <math>10^n</math> is a divisor of <math>209!.</math> Since <math>10=2\cdot5</math> and there are more factors of <math>2</math> than <math>5,</math> finding the number of zeros at the end of <math>209!</math> is the same as finding the largest value of <math>m</math> such that <math>5^m</math> that is a divisor of <math>209!.</math> We then can use the floor function to find the factors of <math>5</math> in <math>209!</math>.This is done by Legendre's formula. Since <math>5^4>209</math>, we only need to compute up to 3. Using the formula, we find that <math>\sum_{i=1}^{3}\lfloor \frac{209}{5^n} \rfloor=\boxed{50}</math>.

Revision as of 18:43, 13 October 2025

Problem

How many zeroes appear at the end of $209$ factorial?

Solution 1

Since we want to find the number of zeros at the end of $209!$, it is the same as finding the largest value of $n$ such that $10^n$ is a divisor of $209!.$ Since $10=2\cdot5$ and there are more factors of $2$ than $5,$ finding the number of zeros at the end of $209!$ is the same as finding the largest value of $m$ such that $5^m$ that is a divisor of $209!.$ We then can use the floor function to find the factors of $5$ in $209!$.This is done by Legendre's formula. Since $5^4>209$, we only need to compute up to 3. Using the formula, we find that $\sum_{i=1}^{3}\lfloor \frac{209}{5^n} \rfloor=\boxed{50}$.

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