Question

How many consecutive zeros would be there at the end 626!-625! ?

Answer 1

626!-625!
=625!(626-1)
=625!*625

In order to form a zero at the end, we need to have a 5 and a 2 in the prime factorization. Since the number of 5's must be smaller than the number of 2's, we only need to consider the number of 5's.

\[\text{Number of 5's in 625!}\]\[=\frac{625}5+\frac{625}{25}+\frac{625}{125}+\frac{625}{625}\]\[=125+25+5+1\]\[=156\]
\[\text{Number of 5's in 625!}\times\text{625}\]\[=156+3\]\[=159\]