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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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Senior SC Moderator
Joined: 14 Nov 2016
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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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12 Feb 2017, 07:12
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95% (hard)

Question Stats:

56% (01:53) correct 45% (02:07) wrong based on 200 sessions

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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. $$2^n - 1$$

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Joined: 02 Aug 2009
Posts: 6802
If n and k are positive integers and (2^n)!/2^k is an odd integer, wha  [#permalink]

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12 Feb 2017, 07:39
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2
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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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.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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13 Feb 2017, 04:23
1
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.
Senior SC Moderator
Joined: 14 Nov 2016
Posts: 1322
Location: Malaysia
Re: If n and k are positive integers and (2^n)!/2^k is an odd integer, wha  [#permalink]

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13 Feb 2017, 04:45
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
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"Be challenged at EVERY MOMENT."

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

Manager
Joined: 03 Jan 2017
Posts: 171
Re: If n and k are positive integers and (2^n)!/2^k is an odd integer, wha  [#permalink]

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18 Mar 2017, 13:50
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
Intern
Joined: 22 Oct 2016
Posts: 22
If n and k are positive integers and (2^n)!/2^k is an odd integer, wha  [#permalink]

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05 Apr 2017, 06:47
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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Sarugiri

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Posts: 8159
Re: If n and k are positive integers and (2^n)!/2^k is an odd integer, wha  [#permalink]

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25 May 2018, 03:49
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Re: If n and k are positive integers and (2^n)!/2^k is an odd integer, wha &nbs [#permalink] 25 May 2018, 03:49
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