# factorial question

• Mar 31st 2010, 09:36 AM
ihavvaquestion
factorial question
determine how many zeros are at the end of the numerals for the following numbers.

a. 10!

b. 100!

c. 1000!
• Mar 31st 2010, 10:28 AM
Soroban
Hello, ihavvaquestion!

Quote:

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.