This is known as the trailing zeros problem.

This function will answer the question for any ,

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- Sep 13th 2008, 02:40 PM #1

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## N!

For a positive integer, , we define N! (read N factorial) to

be the product of all the integers from 1 to N. Thus:

It is clear that N! will end in a zero if N is larger or equal to

5. For big values of N, N! will end in many, many zeros.

How many zeros will end in?

- Sep 13th 2008, 03:16 PM #2

- Sep 13th 2008, 06:35 PM #3

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Hello, perash!

For a positive integer, , we define N! (read N factorial)

to be the product of all the integers from 1 to N.

Thus: .

It is clear that will end in a zero if .

For large values of will end in many, many zeros.

How many zeros will end in?

Plato is absolutely correct . . . Here's the basis of that formula . . .

Every factor of 5 (paired with an even factor) will produce a final zero.

The question becomes: how many 5's are in the prime factorization of ?

Every fifth number has a factor of 5.

. . Hence, there are: . factors of 5.

But every twenty-fifth number has a factor of

. . Each of them contributes*one more 5*to the total.

And there are: . of them.

Hence, contains: factors of 5

Therefore, ends in eighteen zeros.