Shiv2016 wrote:
Bunuel
We calculate trailing zeroes using 5 and not 2. In this, do we assume that the number of two's will be at least equal to the 5's?
Yes, since 2 < 5, then in n! there will be at leas as many 2's as 5's.
Bunuel
We calculate trailing zeroes using 5 and not 2. In this, do we assume that the number of two's will be at least equal to the 5's?
Yes, since 2 < 5, then in n! there will be at leas as many 2's as 5's.
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