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# If n is the product of the integers from 1 to 20 inclusive,

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

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22 Nov 2009, 18:05
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If n is the product of the integers from 1 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
[Reveal] Spoiler: OA
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22 Nov 2009, 18:13
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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22 Nov 2009, 18:25
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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22 Nov 2009, 21:01
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
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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

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Re: If n is the product of the integers from 1 to 20 inclusive,   [#permalink] 20 Apr 2014, 00:49
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