Difference between revisions of "2005 iTest Problems/Problem 37"
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==Solution 1== | ==Solution 1== | ||
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>. | 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>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{iTest box|year=2005|num-b=36|num-a=38}} | ||
| + | |||
| + | [[Category: Introductory Number Theory Problems]] | ||
Latest revision as of 18:44, 13 October 2025
Problem
How many zeroes appear at the end of 
factorial?
Solution 1
Since we want to find the number of zeros at the end of 
, it is the same as finding the largest value of 
such that 
is a divisor of 
Since 
and there are more factors of 
than 
finding the number of zeros at the end of is the same as finding the largest value of 
such that 
that is a divisor of We then can use the floor function to find the factors of 
in .This is done by Legendre's formula. Since 
, we only need to compute up to 3. Using the formula, we find that 
.
See Also
| 2005 iTest (Problems, Answer Key) | ||
| Preceded by: Problem 36 |
Followed by: Problem 38 | |
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