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If n is the product of the integers from1 to 20 inclusive, what is the greatest integer k for which \(2^k\) is a factor of n? A. 10 B. 12 C. 15 D. 18 E. 20

IMO D

1*2 *3 * 4 * ....20

if we take 2 common out of all even numbers from 1 to 20, we have \(2^{18}\). Hence k=18 because \(2^{18}\)is the max factor in form \(2^K\)
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If n is the product of the integers from1 to 20 inclusive, what is the greatest integer k for which \(2^k\) is a factor of n? A. 10 B. 12 C. 15 D. 18 E. 20

\(n=20!\) and we need to find greatest integer \(k\), for which \(2^k*n=20!\). Obviously \(k\) would be highest when \(n=1\). So basically we are asked to determine the highest power of \(2\) in \(20!\).

Re: If n is the product of the integers from 1 to 20 inclusive, [#permalink]

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20 Apr 2014, 00:49

n= 20!

And

2^K * A = n = 20!

=> 20! = 2^K * A

Mathematically , we need to find the powers of 2 contained in the factorial given.

2/4/6/8/10/12/14/16/18/20

Total contribution of '2's' is 18

Hence the answer
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Re: If n is the product of the integers from 1 to 20 inclusive, [#permalink]

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04 Nov 2016, 07:08

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