For any positive integer x, the 2-height of

Sam Kinsman

Hi Nathan,

Again, I'm not sure when we'll be creating a video explanation for this question. But I'm happy to give you a quick explanation!

First, note that what the question describes as 2-height is just the number of 2s in a positive integer x. In other words, take the prime factorization of a number x, and the number of 2s gives you the 2-height. For example, 40 = 2*2*2*5, so it's 2-height is 3.

The question says "is the 2-height of k greater than the 2-height of m?" In other words, it's asking us which number has a greater number of 2s in it: k or m.

Let's look at S1: k > m. This doesn't really help: k could have more or fewer 2s than m does. For example, if k = 7 and m = 6, then m has more 2s. But if k = 8 and m = 7, then k has more 2s. Insufficient.

Now S2: m/k is an even integer. So when m is divided by k, you get an integer. This means that if we take the prime factorization of m and k, we will find that m has all factors of k. And when we divide, the factors get cancelled out and we are left with an even integer. Since we are left with an even integer, we know that it contains a factor of 2, so m has at least one 2 more than k.

Since we know that m has at least one more 2 than k does, we know that the 2-height of m is more than the 2-height of k. So S2 is sufficient, and the answer is B.

Aug 17, 2018 • Reply