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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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Math Expert
Joined: 02 Sep 2009
Posts: 50627
If n and k are positive integers and (2^n)!/2^k is an odd integer, wha  [#permalink]

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19 Feb 2017, 09:07
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Difficulty:

85% (hard)

Question Stats:

52% (02:30) correct 48% (02:19) wrong based on 85 sessions

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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

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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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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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Joined: 04 Jan 2015
Posts: 2203
Re: If n and k are positive integers and (2^n)!/2^k is an odd integer, wha  [#permalink]

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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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19 Feb 2017, 10:44
1
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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Re: If n and k are positive integers and (2^n)!/2^k is an odd integer, wha  [#permalink]

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16 Sep 2018, 23:12
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