1. ## factorial question

determine how many zeros are at the end of the numerals for the following numbers.

a. 10!

b. 100!

c. 1000!

2. Hello, ihavvaquestion!

Determine how many zeros are at the end of the numerals for the following numbers.

. . $(a)\;10! \qquad (b)\;100! \qquad (c)\;1000!$

Every factor of 5 (combined with an even number) creates a final zero.

The question becomes: How many factors-of-5 are contained in $n!$ ?

$(a)\;10!$

We know that: . $10! \:=\:1\cdot2\cdot3\cdots 10$
And every 5th number contains a factor-of-5: . $\left[\frac{10}{5}\right] \:=\:2$ factors-of-5
Therefore: . $10!$ ends in two zeros.

$(b)\;100!$

Every 5th number contains a factor-of-5: . $\left[\frac{100}{5}\right] \:=\:20$ factors-of-5.

But every 25th number contains $5^2$
Each contributes an additional factor-of-5: . $\left[\frac{100}{25}\right]\:=\:4$ more factors-of-5.

Therefore, $100!$ ends in: . $20 + 4 \:=\:24$ zeros.

$(c)\;1000!$

Every fifth number contains a factor-of-5: . $\left[\frac{1000}{5}\right] \:=\:200$ factors-of-5.

But every 25th number contains $5^2$
Each contributes another factor-of-5: . $\left[\frac{1000}{25}\right] \:=\:40$ more factors-of-5.

And every 125th number contains $5^3$
Each contributes another factor-of-5: . $\left[\frac{1000}{125}\right] \:=\:8$ more factors-of-5.

And every 625th number contains $5^4$
Each contributes yet another factor-of-5" . $\left[\frac{1000}{625}\right] \:=\:1$ more factor-of-5.

Therefore, $1000!$ ends in: . $200 + 40 + 8 + 1 \:=\:249$ zeros.