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carsen
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M*) is enoted as the product of all inetgers from 1 to M. If 12 Jul 2004, 07:58
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(*M*) is enoted as the product of all inetgers from 1 to M. If (*24*)/10^n is an integer, and M and N are integers, then what is the maximum value of N.

1.2
2.3
3.4
4.5
5.6

Please explain steps



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M*) is enoted as the product of all inetgers from 1 to M. If 12 Jul 2004, 08:19
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(C) 4

Number: 24!
Numbers which will contribute to the power of 10 are 2, 5, 10, 12, 15, 20.
All these 6 numbers will make maximum x0000.
Hence, (C).
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carsen
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M*) is enoted as the product of all inetgers from 1 to M. If 12 Jul 2004, 08:24
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Thanks JPV ... but it is not clear to me. I undestood the 24!, and why is that the contribution of 2,5,10,20... can u please take time to elaborate. Thanks dude.
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jpv
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M*) is enoted as the product of all inetgers from 1 to M. If 12 Jul 2004, 08:36
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See 24! = 1*2*3*....23*24

We have to find maximum power of 10 which will exactly divide this evenly.

maximum power of 10 means: 10 or 100 or 1000 like that...

If we know how many zeros are present in 24!, then we can find what maximum power of 10 is required.

Now we have to find how many zeros are there in 24!.

For that start looking at individual numbers contributing to 0s in 24!.
U will find these number contributing to 0.

(2*5)*10*(12*15)*20
Look carefully they end with either (2,5) or 0.

See how many 0s u have got in 24!. U can clearly get 4 0s.
Thats what u are looking for. :-D

I guess I have made this clear..
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M*) is enoted as the product of all inetgers from 1 to M. If 12 Jul 2004, 08:38
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Thanks pal. Good explaination. I got the method.
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ian7777
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M*) is enoted as the product of all inetgers from 1 to M. If 12 Jul 2004, 11:58
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great question and great explanation, jpv!

Here's another verion, just to diversify: I love prime numbers. I think they're an excellent way to work out integer problems. Here's how I did this problem.

10 is made up of 2 and 5. Any power of 10 is really a power of 2 and 5. For example, 10^2=(2^2)(5^2).

So the maximum value of N in this case is the maximum number of 2's and 5's that can cancel 24!

Since 24! is made up of, among all the other integers, 5, 10, 15, and 20, and the primes there are 5, 2x5, 3x5, and 2x2x5, we have only 4 fives in 24! There are plenty of 2's, but only 4 fives, so there can only be 4 5's on the denominator of the division. So N's maximum is 4.

I hope that was clear...
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M*) is enoted as the product of all inetgers from 1 to M. If 13 Jul 2004, 02:40
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I was asked the same question (only the numbers ware different) on my admission exam to my hi-school. REALY.



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