[Arithmetic 6] Advanced exercises

Question: How many zeros are there in the last digits of the following number:

$P = 11 \times 12 \times  \ldots  \times 88 \times 89$

        (A) 16         (B) 17         (C) 18         (D) 19         (E) None of the above

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, is simply the multiplicity of the prime factor 5 in n!. This can be determined with this formula: $f(n) = \sum\limits_{i = 1}^k {\left\lfloor {\dfrac{n}{{{5^i}}}} \right\rfloor } = \left\lfloor {\dfrac{n}{5}} \right\rfloor + \left\lfloor {\dfrac{n}{{{5^2}}}} \right\rfloor + \left\lfloor {\dfrac{n}{{{5^3}}}} \right\rfloor + \ldots + \left\lfloor {\dfrac{n}{{{5^k}}}} \right\rfloor$, where k must be chosen such that ${5^{k + 1}} > n$ It's more simple if you look at an example: How many zeros are in the end (after which no other digits follow) of 32! ? According to above 32! has $\left\lfloor {\dfrac{{32}}{5}} \right\rfloor + \left\lfloor {\dfrac{{32}}{{{5^2}}}} \right\rfloor = 6 + 1 = 7$ (denominator must be less than 32, ${5^2} = 32$ is less) So, there are 7 zeros in the end of 32!