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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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New post 12 Feb 2017, 07:12
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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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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, 07:39
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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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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
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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New post 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.
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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, 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.


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

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