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# Number Properties Question

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Number Properties Question [#permalink]  22 Nov 2009, 17:05
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100% (02:04) correct 0% (00:00) wrong based on 3 sessions
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
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Re: Number Properties Question [#permalink]  22 Nov 2009, 17:13
shanewyatt wrote:
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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Re: Number Properties Question [#permalink]  22 Nov 2009, 17:25
Expert's post
shanewyatt wrote:
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!.

Finding highest power of prime in n!: everything-about-factorials-on-the-gmat-85592.html

Hence the power of 2 in 20! would be: \frac{20}{2}+\frac{20}{4}+\frac{20}{8}+\frac{20}{16}=10+5+2+1=18

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Re: Number Properties Question [#permalink]  22 Nov 2009, 20:01
There are 10 numbers divisible by 2
There are 5 numbers divisible by 4
There are 2 numbers divisible by 8
There is 1 number divisible by 16.

Hence the total number of 2’s in 20! are 10+5+2+1=18
Re: Number Properties Question   [#permalink] 22 Nov 2009, 20:01
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