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March 5, 2015

March 5, 2015

Posted by **Joe** on Tuesday, May 31, 2011 at 12:41am.

- Math -
**MathMate**, Tuesday, May 31, 2011 at 7:12amHow many numbers are divisible by 5 between 1 and 20?

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**joe**, Tuesday, May 31, 2011 at 10:18am20,15,10, and 5 are all divisible by 5. so 4. Is that how you solve for the amount of terminal zero's?

- Math -
**joe**, Tuesday, May 31, 2011 at 10:20amuh oh my mistake. I missed a key part of the question. the question is..... How many terminal zeros are at the end of (20!)^2?

- Math -
**MathMate**, Tuesday, May 31, 2011 at 1:23pmThe way we can get a terminal zero in a product depends on two factors (no pun intended), 2 and 5.

There are plenty of two's between 1 and 20, namely we get three of them in 8=2^3, and 16=2^4.

Therefore the governing factor is the number of 5's, which when combined with 2 gives a terminal zero.

If there are four 5's between 1 and 20, then 20! will have 4 terminal zeroes.

As you probably figured it out, squaring doubles the number of terminal zeroes.

So if 20! has 4, then (20!)^2 should have???

Note: watch out if and when you come to 25!

25=5^2, so counts as 2 factors of 5.

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**joe**, Wednesday, June 1, 2011 at 12:34amalright thankyou.

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**MathMate**, Wednesday, June 1, 2011 at 8:07amYou're welcome!

- Math -
**jane**, Friday, June 3, 2011 at 5:05amyou go to sunland dont you?

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