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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 19 Feb 2017, 09:07
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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, 09:32
2^n)!/2^k is an odd integer when n = 2 and k = 3 but this set of numbers are too close for comfort. Next set is n = 3 and k is 7 . 2^7 will take care of all the 2s in 8! making it an odd integer.

Only E fits the bill. 2^3- 1 = 7
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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, 10:28
2
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,
Saquib
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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, 10:44
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1
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


Thanks,
Saquib
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Number Properties – Even Odd | LCM GCD | Statistics-1 | Statistics-2
Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2
Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability
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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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