Bunuel 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.
Kudos for a correct solution.
In simple words, 2-height is just the number of 2s in a positive integer x.
So if x = 40, its 2-height will be 3 because 40 = 8*5 = 2*2*2*5
If x = 15, its 2-height is 0 because there are no 2s in 15.
and so on.
So, an even number will have a 2-height of at least 1.
An odd number will have a 2-height of 0.
"is the 2-height of k greater than the 2-height of m?" means "Does k have more 2s than m?"
(1) k > m
k could have more 2s than m or it could have fewer 2s than m.
For example, if k = 4 and m = 3, k has two 2s while m has none.
If k = 11 and m = 8, k has no 2s while m has 3.
Not sufficient.
(2) m/k is an even integer.
When m is divided by k, you get an integer. So m has all factors of k and they get cancelled out and you are left with an integer. Also, the leftover integer is even so m has at least one 2 more than k.
So 2-height of m is certainly more than the 2-height of k. So we can answer the question that k does not have more 2s than m.
Answer (B)
Thank you very much for this . I didn't even comprehend the question so I certainly had no chance of answering correctly. I was a bit demoralized till I read your great explanation. Does the GMAT use height to represent that amount of times a prime shows up when doing prime factorization often?