The product P(n+1) for n ranging from 5 to 30 means (n+1) takes values from 6 to 31, so we need the number of trailing zeros in the product 6 * 7 * 8 *... * 31 = 31! / 5!. How many trailing zeros does 31!/5! have?

Question

Accepted Answer

None of these. The number of trailing zeros in 31!/5! is determined by the power of 5 (the limiting factor). Powers of 5 in 31! = floor(31/5)+floor(31/25) = 6+1 = 7. Powers of 5 in 5! = floor(5/5) = 1. So 7-1 = 6 trailing zeros. None of the numeric options (111, 147, 137) matches 6, so the answer is 'None of these'.

The product P(n+1) for n ranging from 5 to 30 means (n+1) takes values from 6 to 31, so we need the number of trailing zeros in the product 6 * 7 * 8 ... 31 = 31! / 5!. How many trailing zeros does 31!/5! have?

Solution

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