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How many natural numbers n exist such that the factorial of n concludes with exactly 26 trailing zeros?
- 4
- 5
- 6
- 7
Correct answer: 5
Solution
Trailing zeros Z(n)=floor(n/5)+floor(n/25)+... Z(110)=22+4=26 and this holds for n=110,111,112,113,114; Z(115)=23+4=27. So exactly 5 natural numbers give 26 trailing zeros, which is option index 1, not the stored index 2.
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