Problem 13 · 1991 AJHSME
Hard
Number Theory
trailing-zerosfactor-2-and-5
How many zeros are at the end of the product 25 × 25 × 25 × 25 × 25 × 25 × 25 × 8 × 8 × 8?
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Answer: C — 9.
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Hint 1 of 3
A trailing zero means the number is divisible by 10, and 10 = 2 × 5. So every trailing zero needs ONE 2 paired with ONE 5. Don't multiply this monster out — break each factor into 2's and 5's and count.
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Hint 2 of 3
Each zero eats one 2 and one 5, so the number of zeros is however many COMPLETE 2-and-5 pairs you can form. Count all the 2's and all the 5's; the smaller pile is your answer.
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Hint 3 of 3
Rewrite the pieces as powers: 25 = 5², so 25⁷ = 5¹⁴; and 8 = 2³, so 8³ = 2⁹. Now just compare the two exponents.
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Approach: every trailing zero is one factor of 2 paired with one factor of 5
- A trailing zero comes from a factor of 10 = 2 × 5, so each zero uses up exactly one 2 and one 5. The number of zeros is the number of complete (2, 5) pairs you can build — which is the smaller of (total 2's) and (total 5's).
- Break the product into prime pieces: 25 = 5², so seven 25's give 5¹⁴ (fourteen 5's); 8 = 2³, so three 8's give 2⁹ (nine 2's).
- Fourteen 5's but only nine 2's, so you can form just nine pairs. Trailing zeros = 9.
- Why this transfers: trailing zeros are always limited by the scarcer of factor-2 vs factor-5. In counting zeros of a factorial like 100!, 2's are plentiful, so you just count the 5's — the same min-of-the-two-piles idea.
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