If the expression $15^{6} \times 28^{5} \times 55^{7}$ was evaluated, it would end with a string of consecutive zeros. How many zeros are in this string?
(A) 10
(B) 18
(C) 26
(D) 13
(E) 5
If the expression $15^{6} \times 28^{5} \times 55^{7}$ was evaluated, it would end with a string of consecutive zeros. How many zeros are in this string?
(A) 10
(B) 18
(C) 26
(D) 13
(E) 5

## Solution

A zero at the end of a number results from the product of 2 and 5 . The number of zeros at the end of a number equals the number of product pairs of 2 and 5 that can be formed from the prime factorization of that number.

Since $15^{6} \times 28^{5} \times 55^{7}=(3.5)^{6}\left(2^{2} .7\right)^{5}(5.11)^{7}$

$$
\begin{aligned}
& =3^{6} \cdot 5^{6} \cdot 2^{10} \cdot 7^{5} \cdot 5^{7} \cdot 11^{7} \\
& =3^{6} \cdot 5^{3} \cdot 7^{5} \cdot 11^{7} \cdot 10^{10}
\end{aligned}
$$

There will be ten zeros in the string at the end of the number.

ANSWER: (A)