For any positive integer x, the 2-height of x is defined to be the greatest nonnegative
integer n such that 2^n is a factor of x. If k and m are positive integers, is the 2-height of k
greater than the 2-height of m ?
(1) k > m
(2)m/k is an even integer.
Good one. Took a little more than 2 min to be sure that 1) was not possible.
B for me.
Let the function be TH(x) = n where 2-power-n is a factor of x.
(1) k > m
consider k =12 , m=4 ( TH(k) = TH( m) = 2 in both cases ).
So condition is false
consider k=8, m=4 ( TH(8)= 3 ; TH(4)= 2 ) . So condition is true
So condition is insufficient
(2)if m/k is even there are 2 possibilities.
m and k are odd
m and k are even
But the problem is such that m and k must be even if they must be divisible by 2-power-n.
So with m and k even and m/k being an integer, lets try m=16; k=8
We get TH(x) to be 4 and 3 for m and k.
Try m=20,n=10; Again TH(20) = 2 and TH(10)= 1
So B is sufficient.