How many zeros end 25 factorial in base twelve?

Question

The number 25! (25 factorial, the product of 1 to 25) is written out in base 12.

How many zeros are at the end of it? Explain the method so that it works for any factorial in any base, and say which prime factor turns out to be the bottleneck here and why it is a different one from the familiar base-10 case.

Accepted Answer

Ten zeros. What a trailing zero means. In base 12, each trailing zero is a factor of 12 = 2 squared × 3. So the number of trailing zeros is the largest k such that 12 to the power k divides 25!, which needs 2k factors of 2 and k factors of 3. Counting factors of 2 in 25!. Every second number contributes one factor of 2, every fourth contributes another, and so on. The count is 25/2 + 25/4 + 25/8 + 25/16, each rounded down: 12 + 6 + 3 + 1 = 22 factors of 2. Counting factors of 3. 25/3 + 25/9, rounded down: 8 + 2 = 10 factors of 3. Combine. Each 12 needs two 2s and one 3. The 2s allow up to 22/…

How many zeros end 25 factorial in base twelve?

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