GMAT: Math 43

We're on problem 207. If n is equal to 4p, where p is a prime number greater than 2, how many different positive, even divisors does n have including n? So how many integers go into n, or go into 4p? Well, definitely 4 goes into it, because this is going to be a multiple of 4. Then anything that goes into 4 will also go into it. So 2 goes into it. 2 is also a divisor. p would go into it. Are there any other numbers that go into p? No, p is prime. Well, you know 1 goes into p. But they wanted to know the different positive, even divisors. So, the positive-- we're just dealing in the positive world. That just lets us know this isn't some trick question where we have to count negative numbers. But they want even divisors. So, first of all, 1 is not even. So that can't be it. The second question, is p even? Well, by definition, if p was even, and if it's greater than 2, it would be divisible by 2. So it wouldn't be able to be prime. But they tell us that p is prime. P is prime and it's greater than 2. So it can't be divisible by 2. So it has to be odd. So this is also not an even divisor. So there's only two even divisors, 4 and 2. So the answer is 2. A. Problem 208. S is a set containing 9 different numbers. T is a set containing 8 numbers, all of which are members of S. Which of the following statements cannot be true? The mean of S is equal to the mean of T. So, let's think about this a little bit. So, choice A, that their averages are equal to each other. Let's just say that set S is equal to 1 through 9, 1, 2, 3, 4, 5, 6, 7, 8, 9. So what would its mean be? Well its mean, if you average all these numbers, if you average 1 and 9, you get 5. If you average 2 and 8, you get 5. If you average 3 and 7, you get 5. You average 4 and 6, you get 5. If you average 5, you get 5. So the average is 5. So if this is S, what if T were this? What if T were just this right here? Let me label that. That's T. That's not 7. That's T. Then S is everything. So, it's T plus the 5. So the average of all the numbers in T, if you average 1, 2, 3, 4, 6, 7, 8, 9, your average is going to be 5. If you think about it, the average of 4 and 6 is 5. The average of 3 and 7 is 5. I think you get the idea. You could work it out if you like. If you were to take set S, which is all of these numbers plus the number 5, if you add 5 to a set of numbers whose average is 5, the average is still going to be 5. So A definitely works. B, The median of S is equal to the median of T. Well, once again, that's true with this set. This is a good set to work with. It proves everything we need to prove. What's the median of T. It's the middle number in T. There's 8 numbers, so you have to average the middle two numbers. So, if you average the middle two numbers in T, 4 and 6, you get 5. So 5 is median of T. Then what's the median of S? Well, now S includes the 5. So you can put all these numbers in order. So 5 is the middle number. So 5 is also the median of S. So, both of these can happen, A or B. Choice C, the range of S is equal to the range of T. The range of a set is just the difference between the highest and lowest number. Then the range of S is equal to what? The highest number in S is 9. The lowest number in S is 1. So that equals 9 minus 1, which is equal to 8. What's the range of T? Well the highest number in T is 9. The lowest number is 1. 9 minus 1 is equal to 8. So this one can definitely be true. Statement D, the mean of S is greater than the mean of T. Well, sure. Instead of this being a 5, what if this was a 50? Now all of a sudden the average of T would still be 5. But if you threw this 50 in there for S, your average is going to go a good bit above 5. So statement D could definitely work if you just make that extra number that's in S a lot larger number. Then statement E, which I'm guessing is the choice it cannot be true because we could figure out a case for all of A through D. The range of S is less than the range of T. So the range of the difference between the highest and the lowest. No matter what, the range of S has to be at least as large as the range of T. Think about it. In this case, the range of T is 9 minus 1, which is 8. You could make the range of S larger if you use 50. If 50 was the extra number, then the range is 50 minus 1. But considering that all of these numbers in T are also in S, the difference between the highest number in T and the lowest number in T has to be less than or equal to the range of S. So they're saying the opposite, the range of S is less than the range of T. We know this cannot be true because T is a subset of S. So the choice is E. Next problem-- it took me more time that I wanted to-- 209, How many different positive integers are factors of 441? Well, essentially it's just factoring it. Well let's see, this should be divisible by-- let's see, 4 plus 4 is 8 plus 1 is 9-- so this should be divisible by 3. So 3 goes into 441. 1 times 3 is 1. Bring down a 4, it goes in it 4 times, 12, and then you get 21, 27. So, you get 3 times 147. 3 should also go into this, because 1 plus 4 plus 7 is 12, which is divisible by 3. So 3 goes into 147. 147, you go 4 times 12, 27, 49 times. So let's just do a prime factorization of 441. You get 3 and 147. 147 factors into 3 and 27, which then factors into 3 and 9, which then factors into 3 and 3. OK. So I didn't realize, but this is a [? link ?]. This is the same thing as 3 to the fifth. Is that right? Oh no, sorry. I made a mistake. Let me redo this. 3 goes into 147 not 27 times, but 49 times. So then this is, if you you prime factorization, a 7 and a 7. I was about to say, 441 isn't 3 to the fifth power. So let's write that down here. Let me do it in a different color. I don't want to be confusing. So 441 is equal to 3 times 3 times 7 times 7. So if they want to know how many different positive integers are factors, so all you have to think about is how many positive integers can I construct with two 3's and two 7's? So let's just list them. 3 definitely works. 3 times 3 will definitely work. Well, let's go in order. 7 will definitely work. 3 times 3 will work. That's 9. 3 times 7 would work. 21. 3 times 49 will work. 3 times 7 times 7, so 3 times 49 will work, which was actually 147. 147 works. Then 9 times 7 would work, which is 63. 9 times 7. Then you have 9 times 7 times 7. That's essentially 441. Let me make sure I haven't missed any combinations. You have one 3 and one 7. So you have a 3 and a 7. You have a 3 and two 7's. You have two 3's and a 7. Then you could have two 3's and two 7's. Then, of course, you have the 3 and the 7. So I have 1, 2, 3, 4, 5, 6 numbers. Here I have 1, 2, 3-- what am I missing here? So, I have a 3, 7-- oh, of course. I'm actually missing the 3 times the 3, and I'm missing the 7 times the 7. So, it's 1, 2, 3, 4, 5, 6, 7, 8 factors of which 441-- and that's not one of the choices. How many different positives integers? Oh, and of course, the number 1. So, how many does that come to? 1, 2, 3, 4, 5, 6, 7, 8, 9 numbers. So, that is D. There must be a faster way of doing that. I'll let you think about that one. Anyway. See you in the next video.