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.