goalsnr wrote:

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

OR

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.