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.