SOLUTION: The notation n! = n•(n-1)•(n–2)• • • • • •3•2•1. For example, 5! = 5•4•3•2•1 = 120. How many zeroes occur at the end of the expanded numeral for 100!?

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Question 1115766: The notation n! = n•(n-1)•(n–2)• • • • • •3•2•1. For example, 5! = 5•4•3•2•1 = 120. How many zeroes occur at the end of the expanded numeral for 100!?
Found 2 solutions by ikleyn, MathLover1:
Answer by ikleyn(52834) About Me  (Show Source):
You can put this solution on YOUR website!
.
It was solved and answered under this link
https://www.algebra.com/algebra/homework/word/misc/Miscellaneous_Word_Problems.faq.question.1115588.html

https://www.algebra.com/algebra/homework/word/misc/Miscellaneous_Word_Problems.faq.question.1115588.html



Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
The number of trailing zeros in the decimal representation of n%21, the factorial of a non-negative integer n, can be determined with this formula:
n%2F5%2Bn%2F5%5E2%2Bn%2F5%5E3+...+n%2F5%5Ek where k must be chosen such that 5%5E%28k%2B1%29%3En

so, 100%21 has 100%2F5%2B100%2F5%5E2=100%2F5%2B100%2F25=20%2B4=24 trailing zeros