M12-04

Official Solution:

How many zeros are there at the end of \(100!\) ?

A. 20
B. 24
C. 25
D. 30
E. 32


To find the number of zeros at the end of 100!, we can use this calculation: \(\frac{100}{5}+\frac{100}{5^2}=20+4=24\)

Some background on trailing zeros:

Trailing zeros refer to a sequence of 0's in a decimal representation of a number, after which no other digits are present.

For instance, 125,000 has 3 trailing zeros.

The number of trailing zeros in the factorial of a non-negative integer, \(n!\), can be determined using this formula:

\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where \(k\) must be selected such that \( 5^k \leq n\)

Let's consider an example:

How many zeros are at the end of 32!?

\(\frac{32}{5}+\frac{32}{5^2}=6+1=7\). Observe that the last denominator (\(5^2\)) must be less than 32. Also, note that we only consider the quotient of the division, that is \(\frac{32}{5}=6\).

So, there are 7 zeros at the end of 32!.

Another example: how many trailing zeros does 125! have?

\(\frac{125}{5}+\frac{125}{5^2}+\frac{125}{5^3}=25+5+1=31\),

The formula essentially counts the number of factors of 5 in n!, but since there are at least as many factors of 2, this is equivalent to counting the number of factors of 10, each of which contributes one more trailing zero.


Answer: B