Find the trailing zeros of the number $((100_3)!)$ when it is written in base 3.

The answer is 4.

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In base 3, as with all prime bases, a zero is added to n! only when there is a zero at the end of the number.

100! = 1 * 2 * 10 * 11 * 12 * 20 * 21 * 22 * 100: there are 4 zeros ( 1 from 10, one from 20 and 2 from 100.)