Repetitive Trail
Let denote the number of trailing number of zeros of where is an integer. Which of the following is not a possible value of ?
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1 solution
For every trailing 0, n! must have one factor of 2 and one factor of 5,
note that n! has more factors of 2 than factors of 5,
therefore if n! has a factor of 5 k , it has k trailing zeros.
We can work out, 5! has a factor of 5,
therefore, 25! has a factor of 5^(5x1+1)=5^6
Using this method, we work out that:
125! has a factor of 5^31
and 625! has a factor of 5^156
If n! has 666 trailing 0s, then n! has a factor of 5^666
665=4*156+31+6+4,
Therefore, if m! has 665 zeros, m= 4 625+125+25+4 5=2670
(m+5)! is the first factorial with more than 665 zeros but m+5=2675
2675= 4 625+125+25+5 5 = 4 625+125+2 25
So (m+5)! has 4 156+ 31+ 2 6 =667 0s
Therefore there is no n for which n! has exactly 666 trailing 0s