1991 AMC 8 Problem 13

Attempt Problem 13 of the 1991 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1991 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

13.

How many zeros are at the end of the product 25×25×25×25×25×25×25×8×8×8? \begin{aligned} &25 \times 25 \times 25 \times 25 \\ &\quad {}\times 25 \times 25 \times 25 \\ &\quad {}\times 8 \times 8 \times 8? \end{aligned}
33
66
99
1010
1212
Answer: C
Concepts:trailing zerosprime factorization
Small hint:
A trailing zero comes from a factor of 10=2×510 = 2 \times 5
Big hint:
Count the total factors of 22 and of 5;5; the number of trailing zeros is the smaller count
Solution:
Since 25=52,25 = 5^2, the seven 2525’s give 514.5^{14}. Since 8=23,8 = 2^3, the three 88’s give 29.2^9.
The number of trailing zeros is min⁡(14,9)=9.\min(14, 9) = 9.
Thus, the correct answer is C .
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