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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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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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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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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.
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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, 11:28
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, 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

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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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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17 Sep 2018, 00:12
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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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