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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 \((2^n)!\)= \(\frac{2^n}{2}+\frac{2^{n-1}}{2}+......+\frac{2}{2}\)=\(1+2+2^2+....2^{n-1}=\frac{1*(2^n-1)}{2-1}=2^{n}-1\)
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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
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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 19 Feb 2017, 11:28
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Bunuel wrote:
If n and k are positive integers and (2^n)!/2^k is an odd integer, what is k in terms of n?

A. n
B. 2n−1
C. n^2
D. (n^2+n)/2
E. 2^n − 1


Hey,

PFB the solution. :)

Given:

    • n and k are positive integers.
    • (2^n)!/2^k is an odd integer.

Find:

    • And we need to find out k in terms of n.

Approach:

    • From the second information given to us,
      o \(\frac{(2^n)!}{2^k}\)is an odd integer, we can infer that

      o \(\frac{(1 * 2 * 3 * 4………*2^n)}{2^k}\) is an odd integer

      o This means that all the \(2\)’s in the numerator must be equal to all the \(2\)’s in the denominator.

         If all the \(2\)’s gets canceled, then we will be left with only odd numbers.

      o Therefore, let us find the number of \(2\)’s in the numerator

    • \(\frac{2^n}{2} + \frac{2^{n-1}}{2} + \frac{2^{n-2}}{2}……………+ \frac{2}{2}\)

    • \(2^{n-1} + 2^{n-1} + 2^{n-3}+……………..+1\)

      o The above is a GP series, whose value is \(1*\frac{(2^n-1)}{2-1} = 2^n -1\)

    • Since the number of 2’s in the numerator are \(2^n – 1\), hence we can conclude that

      o \(K = 2^n - 1\)

Hence the correct answer is Option E

Thanks,
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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 19 Feb 2017, 11:44
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Now let us look at another way to solve this.

    • let us take \(n= 2\).

    • We can write the expression as \(\frac{(2^2)!}{2^k}\).

    • Which is equal to \(\frac{4!}{2^k} = \frac{(8*3)}{2^k}\)

    • For the above expression to be an odd integer, the value of k has to be \(3\).

    • Now let us look at the answer choices, put \(n = 2\) and see in which option do we get back \(3\).

      A.\(n = 2\), cannot be the answer.

      B. \(2n−1 = 2 x 2 – 1 = 1\) cannot be the answer

      C. \(n^2 = 2^2 = 4\) cannot be the answer

      D.\(\frac{(n^2+n)}{2} = \frac{(4+2)}{2} = 3\)

      E. \(2^n – 1 = 4 -1 = 3\)

    • So for \(n = 2\), we could eliminate three options. Now we simply need to take \(n =3\).

    • Write the expression as \(\frac{(2^3)!}{2^k}\),

    • Which is equal to \(\frac{8!}{2^k} = \frac{{2*4*6*8*odd numbers}}{2^k}\)

    • For the above expression to be an odd integer, the value of \(k\) has to be \(7\).

    • Plug in \(n = 3\) and check which option gives us \(7\).

    D. \(\frac{(n^2+n)}{2} = \frac{(12)}{2} = 6\) cannot be the answer.

    E. \(2^n – 1 = 8 -1 = 7\). Correct Answer


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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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Re: If n and k are positive integers and (2^n)!/2^k is an odd integer, wha   [#permalink] 17 Sep 2018, 00:12
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