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We're on problem 201 which I had continued in the last video, but, frankly, I had bad reading comprehension. I had read the first statement wrong, and I went down an incorrect path. So let me read it correctly. If the sum of n consecutive integers is 0, which of the following must be true? So I have n numbers in the row, and their sum has to be 0. So, first of all, that tells us that some of them have to be positive and some of them have to be negative. And they're consecutive, right? So if we put this as 0-- let's think about how we can even construct this, how this could ever happen. n consecutive numbers, the sum being 0. Really in, order to do that-- and they're consecutive, which is a trick. If a is one of the numbers, then, really, negative a has to be one of the numbers as well. And why is that? You might say, oh well why can't I have some other set of negative numbers that offset a? Well, the key here is that we're dealing with consecutive numbers. Because those other negative numbers are going to have to offset all of the other positive numbers in between. Maybe I should do this with more abstract. Let's say that n is equal to 1. The sum of n consecutive integers is 0. So, first of all, can n equal 0? Well sure. Only if that one integer is 0. Now what if n is equal to 2? Can n equal 2? Well let's think about it. We have to add one integer to this, and they're integers right? So if we add one integer-- if we add any positive integer here, a, then the sum is going to be 0 plus a. If you're going to cross the negative boundary, if you're going to cross 0, 0 has to be one of your integers. So that's always going to be counted. So, in order to get into negative a, you need to have at least three integers. So, if I said that n is equal to 3, you could do 1, 0, and minus 1. Now, if n is equal to 4, can I just add one number? Can I add either 2 or negative 2? No, that'll change the sum again. So I have to add a 2 and a minus 2. So I think you could see here that there's a pattern because 0 is always going to be one of the integers and it's symmetric on the right- or left-hand side. We're always going to have an odd number of integers we're summing up. So we know that n is odd. So, we know statement one. They say statement one, n even. That's not true. We actually can't construct it with n as even. Statement two says, n is odd which we've just shown. And then statement three is, the average of the n integers is 0. Well sure. What's the average? It's the sum of the integers divided by n. Well, they're telling us that the sum of the integers is 0. So 0 divided by anything, as long as we don't have zero integers, is 0. So two and three are both correct. So the choice is e. Next question. 202. In the formula, v is equal to 1 over 2r cubed. If r is halved, then v is multiplied by what? The easiest thing I always find to do in this is take numbers. So let's say, if r is 2 what happens? And let's halve it when r equals 1 and see what happens to v. We can do it analytically, but this is, frankly, the easiest way to do it. So you can say when r is equal to 2, v is equal to 1 over 2 times 2, is 4, to the third power is 64. And then when r is 1, you get 1 over-- let's see, 2 times 1 is 2 to the third power. So you get 1/8. So what happened to v? We, essentially, multiplied that by 8. So that is choice b. To go from 1/64 to 1/8 you have to multiply by 8. You can view 1/8 as the same thing as 8/64 if that clarifies it at all. I always find it easier to deal with actual physical numbers where they say, if you change this variable by a factor of 5 what happened to that variable? So easy to just-- well anyway. It could get very confusing if you try to do an r1 and an r2, and r1 is 2 times r2 and all that. Next question. 203. A certain bakery has six employees. It pays annual salaries of $14,000 to each of two employees. $16,000 to one employee. And $17,000 to each of the remaining three employees. The average annual salary of these employees is closest to which of the following? So, we have to take the sum of all their salaries and divide by 6. So it'll be 2 times 14, plus 1 times 16-- which is 16, plus 3 times 17, and we divide it by 6. 2 times 14 is 28, plus 16, plus 51, all of that over 6. Let's see, what is that equal to? 28, 16, 51, 8 plus 6 is 14, 14 plus 1 is 15, 1 plus 2 is 3, 3 plus 1 is 4, 4 plus 5 is 9. This was 8 plus 6. So, it's 95 divided by 6. Sorry, 95 divided by 6. I think my brain was computing ahead of me because 96 divided by 6 is equal to 16. So this number is going to be just under 16. It's 15 and 5/6. So they want to get an approximate, as close as-- oh, actually they have a lot of numbers that are really close. Let me make sure that I get it right. So, 6 goes into 95-- if they didn't have numbers that weren't close, I wouldn't do this. 6 goes into 95, 6 into 9, 6. 35, goes 5 times, 5 times 6 is 30, remainder of 5. Actually let's keep going because we want a decimal. We're doing it in thousands So, bring down a 0. 6 goes into 50 8 times. 8 times 6 is 48. Remember, we were dealing in thousands when I said 16, that was thousands. So, now we're at 15.8 thousand. So 8 times 48, 20, 6 goes into 20 3 times. I think that's close enough to what we need. So it's going to be roughly $15,800. Or 15.8 thousands. And that is choice C. At first I said maybe we can approximate, and they had 16,000 there, but then I saw the 15,800 so we should do a little bit more math. We don't want to get too careless. Next problem. 204. If x is equal to the sum of the even integers from 40 to 60 inclusive. This is my ad hoc notation. And y is the number of even integers. Let's see, if x is equal to the sum of the even integers from 40 to 60, inclusive, so it includes 40 and 60, and y is the number of even integers from 40 to 60 inclusive, what is the value of x plus y? The number-- y is an easy thing. How many integers are there between 40 and 60, if you include both of them? This is always a tricky thing to think about. But, if you think about it, if I said the numbers 1, 2, and 3, how many integers are there here? Well, clearly, you can count them. 1, 2, and 3. But if you wanted a quick way you could say 3 minus-- if I just took the endpoint minus the starting point, I would get these integers. Then I want to add 1. If I want to know how integers are there between 60 and 40, I would say 60 minus 40. And that'll give me 41, 42, 43, 44, all the way to 59, and 60. And then I have to add 1 because we're including 40 as well. So y is equal to 21. Now let's think about, is there an easy way to add up all of the integers between 40 and-- all of the even, oh no, no sorry. y is the number of even integers from 40 to 60. So that's not right. So, even integers. Frankly, it might just be easier to-- let's just write them out. 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, and 60. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. 11 intergers. So that's y. And we need to take the sum of all of these. So there's a couple of ways we could think about it. You could view this as 40 plus 0, 40 plus 2, 40 plus 3, and so forth. All the way to 40 plus 8. And so if you wanted to add up at least up to this point, it's a pretty straightforward way of doing it. You could just say, there's 1, 2, 3, 4, 5 of them. So it's equal to 40 times 5. That adds up all the 40's. Plus 2, plus 4, plus 6, plus 8. So 2 plus 4 is 6, plus 6 is 12, plus 8 is 20. So, plus 20. So, that's 40 times 5 is what? 200 plus 20 is equal to 220. And you use the same logic here. This is 50 plus 0, 50 plus 2, all the way to 50 plus 10. And how many are there? There's 1, 2, 3, 4, 5, 6. So, it's equal to 6 times 50, plus 2 plus 4 is 6, plus 6 is 12, plus 8 is 20. Which we actually already knew. We already calculated it there. It was 20. Because these don't contribute to the-- oh no, no plus 10. Sorry, plus 10. Don't want to get careless. Plus 10. Because there's six of these. There's plus 0, plus 2, plus 4, plus 6, plus 8, plus 10. So, it's 20 plus 10 is 30. 2 plus 4 plus 6 plus 8 plus 10 is 30. So, 6 times 6 is 300, plus 30 is equal to 330. So, 330 plus 220. These numbers have a strange pattern that-- well there is a formula if you want to add the first n numbers, but they've changed it a little bit by talking about even numbers so I wanted to do it manually. So you go 0, 5, 5. So 550 is the sum. And then we know that y-- this is x. x is equal to 550. And y is equal to 11. We figured that out before. So x plus y is equal to 550, plus 11 which is equal to 561, which is, thank goodness, actually one of the choices. That's choice D. See you in the next video.