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VietDao29
#3
Aug4-05, 07:02 AM
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Quote Quote by SteveRives
Your question is the same as finding out how often 5 is a prime factor on the natural numbers.

So I am guessing that 100! has 20 zeros, because in counting by fives, we get to 100 after 20 steps.
You are WRONG, SteveRives.
24 zeros is the answer.
(*)You will have a number which is divisible by 5 for every 5 successive integer.
(*)You will have a number which is divisible by 25 (ie 5.5) for every 25 successive integer.
(*)You will have a number which is divisible by 125 (ie 5.5.5) for every 125 successive integer, and so on...
So you will have 20 numbers which is divisible by 5 (included numbers that is divisible by 25).
And you will have 4 numbers which is divisible by 25.
So you will have: 20 + 4 = 24 zeros in 100!
In general, you will have: n! has:
[tex]C = \left[ \frac{n}{5} \right] + \left[ \frac{n}{5 ^ 2} \right] + \left[ \frac{n}{5 ^ 3} \right] + ... + \left[ \frac{n}{5 ^ q} \right][/tex] zeros at the end.
You can stop the sum at any q such that:
[tex]\left[ \frac{n}{5 ^ q} \right] = 0[/tex]
Where [...] denotes the integer part of a number. For example : [1.55] = 1, [0.77] = 0, [14.333333] = 14.
Viet Dao,