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a difficult prob, pls help to solve

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by tracyyahoo » Sat Sep 03, 2011 3:43 am
Pls help to explain in detail, I need explainations.

tracyyahoo wrote:If n is the greatest positive integer for which 2^n is a factor of 10!, then n =
A. 2 B. 4 C. 6 D. 8 E. 10
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by Krk » Sat Sep 03, 2011 6:00 am
My way of approach to this problem.
10! = 1*2*3*4*5*6*7*8*9*10
This can be re-written as:

10! = 1*2*3* (2*2) * 5 * (2*3) * 7 * (2*2*2) * (3*3) * (2*5)
rearranging
10! = 2* (2*2) * (2) * (2*2*2) * (2) * 3 * 5 * (3) * 7 * (3*3) * 5
10! = 2^8 * 3*3*3*3 * 5*5 * 7

(3*3*3*3 * 5*5 * 7) is not divisible by 2.
Hence,
10! is factor of 2^8
Hence, the greatest number n is 8
Answer is option D
Please reply your feedback on the post. Thanks
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Krk
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by Abhishek009 » Sat Sep 03, 2011 6:41 am
tracyyahoo wrote:If n is the greatest positive integer for which 2^n is a factor of 10!, then n =
A. 2 B. 4 C. 6 D. 8 E. 10
10 ! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

or 10! = (2*5) * (3*3) * (2*2*2) * 7 * (2*3) * 5 * (2*2) *3 *2 * 1


Now find the powers of 2 in 10!

There are 8 2's in 10 ! , so the highest power of 2 in 10! is 8....
Abhishek
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