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Question on terminating zero's?

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Question on terminating zero's? [#permalink] New post 16 Jun 2013, 01:21
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If we were given a question to find the number of zero's in the product of the first 100 integers then can we use the following

we know that to find the zero's we need 5 and 2 so we can focus on the 5's

100/5 + 100/ 5^2 + 100/5^3 + 100/5^4....

we get 20 + 4 + 0 + 0 ( I know you can stop at 5^3 but just wrote the extra step for a general case)

so there will be 24 zeros? Is this approach correct and can I use it for all such questions
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Re: Question on terminating zero's? [#permalink] New post 16 Jun 2013, 01:28
Yes, that's correct.

Whenever you are asked for the trailing zeros of a given integer, just find the total number of 2-5 pairs.

So total trailing zeros of 43! is 9 for example:
40*...*35*...*30*..*25*...*20*...*15*..*10*..*5 they have 9 fives (25 has 2) so there are 9 zeros.

Most of the times it's all about counting the 5s because the number of 2s will be greater
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Re: Question on terminating zero's? [#permalink] New post 16 Jun 2013, 02:10
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fozzzy wrote:
If we were given a question to find the number of zero's in the product of the first 100 integers then can we use the following

we know that to find the zero's we need 5 and 2 so we can focus on the 5's

100/5 + 100/ 5^2 + 100/5^3 + 100/5^4....

we get 20 + 4 + 0 + 0 ( I know you can stop at 5^3 but just wrote the extra step for a general case)

so there will be 24 zeros? Is this approach correct and can I use it for all such questions


Trailing zeros:
Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

125000 has 3 trailing zeros;

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:

\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}, where k must be chosen such that 5^(k+1)>n

It's more simple if you look at an example:

How many zeros are in the end (after which no other digits follow) of 32!?
\frac{32}{5}+\frac{32}{5^2}=6+1=7 (denominator must be less than 32, 5^2=25 is less)

So there are 7 zeros in the end of 32!

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.

BACK TO THE ORIGINAL QUESTION:

According to above 100! has \frac{100}{5}+\frac{100}{25}=20+4=24 trailing zeros.

For more on trailing zeros check: everything-about-factorials-on-the-gmat-85592-40.html

Similar questions to practice:
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Hope it helps.
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Re: Question on terminating zero's?   [#permalink] 16 Jun 2013, 02:10
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