2007 Cyprus MO/Lyceum/Problem 28

Problem

The product of $15^8\cdot28^6\cdot5^{11}$ is an integer number whose last digits are zeros. How many zeros are there?

$\mathrm{(A) \ } 6\qquad \mathrm{(B) \ } 8\qquad \mathrm{(C) \ } 11\qquad \mathrm{(D) \ } 12\qquad \mathrm{(E) \ } 19$

The number of zeros at the end of a number is determined by the number of 2's and 5's in its prime factorization.

$15^8\cdot28^6\cdot5^{11}=2^{12}\cdot3^8\cdot5^{19}\cdot7^6$

There are 12 $2$ 's and 19 $5$ 's. Each pair adds a zero, but any extras don't count (in this case, the 7 extra $5$ 's).

$12\Longrightarrow\mathrm{ D}$