If you're seeing this message, it means we're having trouble loading external resources on our website.

Jeżeli jesteś za filtrem sieci web, prosimy, upewnij się, że domeny ***.kastatic.org** i ***.kasandbox.org** są odblokowane.

Główna zawartość

Aktualny czas:0:00Całkowity czas trwania:12:38

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.