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?

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

Solution

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).

$12\Longrightarrow\mathrm{ D}$

See also

2007 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 27
Followed by
Problem 29
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