GMAT: Data sufficiency 38

We're on problem 145. It says, is 1/p greater than r/r squared plus 2. That's their question. Statement number one-- I realize I go on these tangents before even looking at the statement; I should look at the statements first-- p is equal to r. So let's see if we can simplify this. So if p is equal to r, then we get 1/r-- instead of a p, right?-- is greater than r/r squared plus 2. Let's multiply both sides of this equation by r. Well, we don't know that r is necessarily greater than 0. That's the problem. So if r is greater than 0-- well, it's going to change the inequality one way or the other. So let's just assume r is greater than 0. So if we multiply both sides of the equation by r, then we don't have to switch the inequality, because we're assuming r greater than 0. That's an assumption. Then you get 1 is greater than r squared over r squared plus 2. And then if r is greater than 0, then r squared plus 2-- actually, r squared plus 2 is always going to be greater than 0, because r squared [INAUDIBLE] positive. Let's multiply both sides of this equation by r squared plus 2. And you get r squared plus 2 is greater than r squared. You could subtract r squared from both sides, and you get 2 is greater than 0, which is true. But remember, I had to make this assumption that r is greater than 0. If you assume that r less than 0, then all of this is going to break down. Because when you multiply both sides by r, you have to switch the inequality to get 1 is less than r squared over r squared plus 2. And then, eventually, you're going to end up with r squared plus 2 is less than r squared. Or that 2 his less than 0, which is false. So this statement alone isn't enough. We have to know whether or not r is greater 0. Statement two-- well, there you go; I really should look at the statements first-- r is greater 0. And just so you know, statement two by itself is insufficient, because if you know that r is greater than 0, you have no idea what p is. So you can't answer it yet. But both statements combined are sufficient to answer this question. Next question. 146. Is n an integer? Statement one says n squared is an integer. Well, that doesn't help us. I mean, what if n was equal to the square root of 2? And then n squared would be equal to 2, which is an integer. But this is clearly not an integer. But on the other hand, n could be equal to 2, in which case n squared would be an integer. So whether or not n squared is an integer, both of these are cases where n squared are integers, but one ends up where n is an integer, one is where n isn't an integer. So this, by itself, is not enough to tell me whether n is an integer. Statement two tells us, the square root of n is an integer. OK. So that essentially tells us that n is equal to some integer squared, right? You could take the square root of both sides. You'd get the square root of n is equal to some integer. So you take any integer, you square it, you're going to get an integer. So statement two alone is sufficient to answer this question. That was a strangely easy question. They they were getting a little bit hairy and confusing, but that was a nice little rest question, I think. 147. If n is a positive integer-- so n is a positive integer. Well, n is greater than 0. It's an integer. Well, I didn't write that down. Is n to the third minus n divisible by 4? Fascinating. Let's see. So my initial reaction is to see if I can simplify this as a product of simpler expressions. I look already at statement number one. Let me look at statement number one. Statement number one says, n is equal to 2k plus 1, where k is an integer. I don't feel like taking 2k plus 1 and cubing it. That's not an easy thing to do. It'll take some time. So my intuition is that maybe we should simplify this a little bit. So if we factor out an n, that's equal to n times n squared, minus 1. And that's equal to n times-- what's n squared minus 1? That's n plus 1, times n minus 1. And now this is something that's much easier to substitute this into. So let's do that. Let's substitute n equals 2k plus 1 into this. So if n is equal to 2k plus 1, you get 2k plus 1, times 2k plus 1, plus 1. So that's 2k plus 2. That's this one. And then you have 2k plus 1, minus 1. So that's just 2k. And then what does that simplify to? You get 2k plus 1. Remember, in the back my mind, I want a 4 to show up, because I want this thing to be divisible by 4. 2k plus 1, times 4k squared plus 4k. Interesting. So now I can factor out a 4. So it's 2k plus 1, times k squared plus k, times 4. And we know that this is an integer, because they told us that k is an integer. We know that this is an integer, because k is an integer. And we know 4 is an integer. So we've essentially kind of factored n cubed minus n. And 4 is one of it's factors. So it's definitely divisible by 4. I mean, we can divide it by 4 right now. If we divide it by 4, we'd be left with this, which is clearly an integer, because k is an integer. So n to the third minus n is divisible by 4 if we can assume that n is equal to 2k plus 1, where k is some integer. Now, what do they tell us in statement two? Statement two tells us, n squared plus n is divisible by 6. Well, I don't see how that's helpful at all. n squared plus n is divisible by 6. You can't even relate that to that. I mean, n squared-- I could try to do something fancy here. No. This is so different than this. n squared plus n is so different from n to the third. I could factor out an n. I could say n times n plus 1 is divisible by 6. Well, I guess that's interesting. That tells us that this part is divisible by 6. n times n plus 1 is divisible by 6. Do they tell us that n is a positive integer? Yeah, they do tell us that n is a positive integer. So this could be 2 and 3. If this is 2 and 3, then this would be 1. This could be 2 and 3. In fact-- well, this could be 2 and 3, but it could also be-- well, actually, that's the only numbers it could be. It could be 2 and 3, in which case this is 1, in which case this whole thing-- if you say this, then n could be 2. n is equal to 2. n plus 1 is equal to 3. And then n minus 1 would be equal to 1. And let me see. Is that the only case? I'm looking at the answer right now, just because I want to make sure. They say that this isn't sufficient. But, once again, kind of like problem 142, actually-- because if you say that n is a positive integer and that n times n plus 1 is divisible by 6, I can't think of any other numbers where you multiply one number times the next number, and their integers, where you get a multiple of 6. No, I take that back. It could be 3 and 4. There could be 3 and 4. It could be-- yeah, there's actually a bunch of them. This could be 3, this could be 4, because it's a multiple 6, and then this would be 2. So that's right. This is not enough information. So that's good. Because by this information, n could be 2, n plus 1 could be 3, in which case, this number would be actually 6. But that still doesn't help us, because 6 isn't divisible by 4. But then I could come up with the situation where this number right here, where n is 3, n plus 1 is 4, and n minus 1 is 2, where this does become divisible by 4. So this doesn't give us enough information. Not useful. And I'm out of time. I'll see you in the next video.