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Current time:0:00Total duration:8:32

We're on problem 248, truly
the home stretch. Right triangle pqr is to be
constructed on the xy plane. Let me draw an xy plane,
that seems like it should be useful. So that's x, and y -- I'm sorry,
that's y and that's x. Let me label them since
I said them wrong. y and x, that should be second
nature to you at this point. If you start the xy plane so the
right angle is at p, and pr is parallel to the axis
-- so it's going to look something like that, where
that's p and that's r, parallel to-- and p is the right
angle, right, so it's going to look something
like that. So q has to be right below p. The x- and y-coordinates of p,
q, and r are to be integers that satisfy the inequality
minus 4 is less than or equal to x, which is less than
or equal to 5. And it tells us that 6 is less
than or equal to y, which is less than or equal to 16. How many different triangles
with these properties could be constructed? So let's first draw
this range. So x could be between
minus 4 and 5. And x and y are always
going to be integers, so that's our boundary. And it could be minus
4 or 5, right? And then y is between
6 -- let's just say this is 6 -- and 16. So let's think about this. Let's think about where we
can place the point. So first of all, how many
different legitimate places can we put the point p? Let's just kind of figure out
how many different ways we can construct this thing. So if we have to pick its
x-coordinate, how many different x-coordinates
are valid? So you have minus 4, minus 3,
minus 2, minus 1, 0, 1, 2, 3, 4, and 5. Right? They should be -- 5 minus
5, so it should be 9. 1, 2, 3, 4, 5, 6, 7, 8, 9 --
oh actually, no, it's 10 different points, because we
could include minus 4, right? So it could be -- x could be
minus 4, minus 2, 3 -- right, this is minus 3, minus 2, minus
1, 0, 1, 2, 3, 4, and 5. Right? So this x value can take on
any of these 10 spots. So p can take on 10
different values. Well, its x-value could take on
10 different values, right? So p's x can take on 10
different values. And how many values can
its y take over? Well, this top boundary is 16,
this bottom boundary is 6. And we can be either
one, right? So 16 minus 6 would be 7, 8, 9,
10, 11, 12, 13, 14, 15, 16. But we can include
6 as well, right? If you just said 16 minus 6,
that would give you 10 spots, but since we're including 6, the
y could take on 11 spots. You could count them, maybe
it'll be beneficial for you to see all the different
y's it could be. It could be anything between 6
and 16, including 6 and 16. So there's 11 possibilities
for the y. So that means that there are
110 different possible p's. Let me just draw one of them. Let's say that's our p. So our p is there. For any given p, how
many different possible r's are there? Well r is in the
same line as p. So whatever p's y value is,
r has to be the same. But r could be any of the other leftover x spaces, right? So if p took up one of them, r
could be here, it could be here, it could be here. r could be in any of
these spots, right? So how many left over
x spots are there? Well there were 10 x spots,
10 x-coordinates. p took up one of them. So now r could only be in one
of nine x-coordinates. Fair enough. And r is limited to its y. Once we pick a y coordinate for
p, that's got to be the same coordinate as r, right?
r has to be on the same y-coordinate as p. Because they told us
that this line is parallel to the x-axis. Fair enough. So how many possibilities --
once we've picked a p and an r, how many different
possibilities do we have for q? Well q is going to have the
same x value, because they told us that this is a
right angle, and so this has to be q. So q has to be in the same
vertical line as p. It has to have the
same x-value. And so on how many different
y-values can it have? Well there were a total of -- we
said there were originally 11 potential y-values, right? That's what we said when we
picked our p. p could be anywhere between 6 and 16. q could be any of those numbers,
but it can't be at the same point as p, right? This side has to have
some length. So if there were 11 original
slots, 11 original y slots, p takes one of them up. So then q could be any
of one of 10 y slots. So if we want to know the total
number of potential triangles like this, there are
110 potential places to put p. Once we've put p, there are nine
potential places to put r, and once we've put p, there's
also 10 potential places to put q. So this is what, 110 times 90. That's the same thing as 11
times 9 with two 0's, right? Which is choice c. And we're onto our last -- I'm
glad this a new video, that took me six minutes. Problem 249. Hopefully it won't take you
that long on the GMAT. It takes less time if you don't
have to explain it. If n is a positive integer less
than 200 -- so n is less than 200, it's positive and it's
an intiger -- and 14 over n -- 14n over 60
is an integer. So that means that 14n divided
by 60 -- or 14 is divisible by 60 then n has how many different
prime factors? Let's simplify this first.
Divide the top and the bottom by 2. You get -- this is also saying
that 7n over 30 is an integer. Well multiplying by 7 doesn't
make it any easier to be divisible by 30, right? Because 7 and 30 share
no common factors. 14 and 60 did, 2, but
we got rid of it. So we have 17 and 30, so n has
to be divisible by 30. Right? n has to be divisible
by 30 for this to be an integer. Well, I don't know, let's just
pick the smallest one. I mean, there's a lot of
different candidates. n could be 30, or 60,
or 90, or a bunch of different numbers. But let's just pick
n is equal to 30. They just say that n is less
than 200, so let's just say that n is equal to 30. If n is equal to 30, then they
say, how many different positive prime factors? Well let's just do a prime
factorization of 30. 2 times 15, which
is 3 times 5. So it has 3 positive
prime factors. So that's choice b. I don't know, I'm
just curious. So let's see if that would've
worked if we picked n equals 60. If you'd have done the prime
factorization of 60, you'd have gotten 2 times 30,
which is 2 times 15, which is 3 times 5. And yeah, you still have
only three factors. You're using one
of them twice. So this is the same thing as
2 squared times 3 times 5. So you still have actually
three different positive prime factors. So it would work for any
of the multiples of 30. Anyway, choice c. And we're all done. I don't know what I'm going to
do next, maybe I'll do some LSAT videos, or maybe
I'll get back to actually teaching something. Anyway, see you in a
future play list.