posted by Jon on .
What is the smallest number n such that n! ends in at least 1000 zeros in base 10 notation
you need 10 to be a factor 100 times.
That means you need 2^100 and 5^100 to be factors.
If [n] is the greatest integer <= n,
n! has 5 as a factor [n/5]+[n/25]+[n/125]+... times.
So, solve [n/5]+[n/25]+[n/125]+[n/625]+[n/3125] = 1000.
= 801+160+32+6+1 = 1000
Naturally, 2 occurs as a factor many more times than this, so 4005! is the first factorial to end in 1000 zeros.