determine how many zeros are at the end of the numerals for the following numbers.
a. 10!
b. 100!
c. 1000!
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Determine how many zeros are at the end of the numerals for the following numbers.
. . $\displaystyle (a)\;10! \qquad (b)\;100! \qquad (c)\;1000!$
Every factor of 5 (combined with an even number) creates a final zero.
The question becomes: How many factors-of-5 are contained in $\displaystyle n!$ ?
$\displaystyle (a)\;10!$
We know that: .$\displaystyle 10! \:=\:1\cdot2\cdot3\cdots 10$
And every 5th number contains a factor-of-5: .$\displaystyle \left[\frac{10}{5}\right] \:=\:2$ factors-of-5
Therefore: .$\displaystyle 10!$ ends in two zeros.
$\displaystyle (b)\;100!$
Every 5th number contains a factor-of-5: .$\displaystyle \left[\frac{100}{5}\right] \:=\:20$ factors-of-5.
But every 25th number contains $\displaystyle 5^2$
Each contributes an additional factor-of-5: .$\displaystyle \left[\frac{100}{25}\right]\:=\:4$ more factors-of-5.
Therefore, $\displaystyle 100!$ ends in: .$\displaystyle 20 + 4 \:=\:24$ zeros.
$\displaystyle (c)\;1000!$
Every fifth number contains a factor-of-5: .$\displaystyle \left[\frac{1000}{5}\right] \:=\:200$ factors-of-5.
But every 25th number contains $\displaystyle 5^2$
Each contributes another factor-of-5: .$\displaystyle \left[\frac{1000}{25}\right] \:=\:40$ more factors-of-5.
And every 125th number contains $\displaystyle 5^3$
Each contributes another factor-of-5: .$\displaystyle \left[\frac{1000}{125}\right] \:=\:8$ more factors-of-5.
And every 625th number contains $\displaystyle 5^4$
Each contributes yet another factor-of-5" .$\displaystyle \left[\frac{1000}{625}\right] \:=\:1$ more factor-of-5.
Therefore, $\displaystyle 1000!$ ends in: .$\displaystyle 200 + 40 + 8 + 1 \:=\:249$ zeros.