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2007 Cyprus MO/Lyceum/Problem 28

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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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.
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