If n and k are positive integers and (2^n)!/2^k is an odd integer, wha

Question

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

chetan2u · Accepted Answer

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

sobby · Answer

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.

EgmatQuantExpert · Answer

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 Quant Expert e-GMAT Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts

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EgmatQuantExpert · Answer

Now let us look at another way to solve this.

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

E. \(2^n – 1 = 4 -1 = 3\)

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

Thanks,
Saquib
Quant Expert
e-GMAT

Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts

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sarugiri · Answer

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.

bumpbot · Answer

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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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