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If n and k are positive integers and (2^n)!/2^k is an odd integer, wha

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If n and k are positive integers and (2^n)!/2^k is an odd integer, wha [#permalink]

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If n and k are positive integers and \(\frac{(2^n)!}{2^k}\) is an odd integer, what is \(k\) in term of \(n\)?

A. \(n\)

B. \(2n - 1\)

C. \(n^2\)

D. \(\frac{(n^2 + n)}{2}\)

E. \(2^n - 1\)
[Reveal] Spoiler: OA

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If n and k are positive integers and (2^n)!/2^k is an odd integer, wha [#permalink]

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New post 12 Feb 2017, 06:39
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ziyuenlau wrote:
If n and k are positive integers and \(\frac{(2^n)!}{2^k}\) is an odd integer, what is \(k\) in term of \(n\)?

A. \(n\)

B. \(2n - 1\)

C. \(n^2\)

D. \(\frac{(n^2 + n)}{2}\)

E. \(2^n - 1\)



Hi,

Two ways..

1) substitution

If you do not know the proper method..
Let n =3
\(\frac{(2^n)!}{2^k}\)
Then 2^3=8 and number of 2s in 8!=8/2 + 8/4 + 8/8=4+2+1=7..
See which choice gives you 7 as answer.

A. \(n\)....=3 No
B. \(2n - 1\)....2*3-1=5 No
C. \(n^2\)....2^3=8 No
D. \(\frac{(n^2 + n)}{2}\)..(2^2+2)/2=3 No
E. \(2^n - 1\) 2^3-1=8-1=7... ANSWER

E

2) proper method
Basically number of 2s in \((2^n)!\)= \(\frac{2^n}{2}+\frac{2^{n-1}}{2}+......+\frac{2}{2}\)=\(1+2+2^2+....2^{n-1}=\frac{1*(2^n-1)}{2-1}=2^{n}-1\)
E
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html


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Re: If n and k are positive integers and (2^n)!/2^k is an odd integer, wha [#permalink]

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New post 13 Feb 2017, 03:23
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ziyuenlau wrote:
If n and k are positive integers and \(\frac{(2^n)!}{2^k}\) is an odd integer, what is \(k\) in term of \(n\)?

A. \(n\)

B. \(2n - 1\)

C. \(n^2\)

D. \(\frac{(n^2 + n)}{2}\)

E. \(2^n - 1\)


Checking with answers choice is the easy method , ( it will not drain the brain too much :p )

My approach : finding a pattern

let n = 2

4!/2^k = 4*3*2*1 /2^k
for odd number we need to cancel out each number having 2 as a factor..so we need to cancel out 4 and 2
i'e K = 3

lets check for n= 3
8!/2^k = 8 * 7 * 6 * 5 *4 *3 *2*1 , for cancelling each number having factor 2 we require k = 7

same we will test for n = 4
we will need k = 15

we got a pattern as

n=2 / k = 3
n=3 /k = 7
n=4 / k =15

so solution will be 2^n - 1.
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Re: If n and k are positive integers and (2^n)!/2^k is an odd integer, wha [#permalink]

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New post 13 Feb 2017, 03:45
sobby wrote:
ziyuenlau wrote:
If n and k are positive integers and \(\frac{(2^n)!}{2^k}\) is an odd integer, what is \(k\) in term of \(n\)?

A. \(n\)

B. \(2n - 1\)

C. \(n^2\)

D. \(\frac{(n^2 + n)}{2}\)

E. \(2^n - 1\)


Checking with answers choice is the easy method , ( it will not drain the brain too much :p )

My approach : finding a pattern

let n = 2

4!/2^k = 4*3*2*1 /2^k
for odd number we need to cancel out each number having 2 as a factor..so we need to cancel out 4 and 2
i'e K = 3

lets check for n= 3
8!/2^k = 8 * 7 * 6 * 5 *4 *3 *2*1 , for cancelling each number having factor 2 we require k = 7

same we will test for n = 4
we will need k = 15

we got a pattern as

n=2 / k = 3
n=3 /k = 7
n=4 / k =15

so solution will be 2^n - 1.


Thank you for your approach. It helps me to understand better.


Sent from my iPhone using GMAT Club Forum mobile app
_________________

"Be challenged at EVERY MOMENT."

“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”

"Each stage of the journey is crucial to attaining new heights of knowledge."

Rules for posting in verbal forum | Please DO NOT post short answer in your post!

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Re: If n and k are positive integers and (2^n)!/2^k is an odd integer, wha [#permalink]

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New post 18 Mar 2017, 12:50
I tried smart numbers: let's n=2, then K is 3, too many answers fit
then I tried n=3, then K was 7, E is left
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If n and k are positive integers and (2^n)!/2^k is an odd integer, wha [#permalink]

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New post 05 Apr 2017, 05:47
This can be done easily by the properties of factorial and the power of prime p in n!
n! = n * (n-1)!

n/p + n/p^2 + n/p^3 + ... + n/p^k, where k must be chosen such that p^k <=n
Refered from : https://gmatclub.com/forum/everything-a ... 85592.html

Question is asking to find power of 2 in (2^n)! which is (2^n)!/2^k
(2^n)! = 2^n * (2^n-1)!
(2^3)! = 2^3 * (2^3 -1)! = 8 * 7!
For example, (2^3)!/2^k = 8!/2 = 8/2 + 8/4 + 8/8 = 4+2+1 = 7, which is satisfied only by (2^3 -1). Hence option E
Correct me if i'm wrong.

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If n and k are positive integers and (2^n)!/2^k is an odd integer, wha   [#permalink] 05 Apr 2017, 05:47
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