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Current time:0:00Total duration:13:51

We're on problem 151. If x and y are positive,
is the ratio of x to y greater than 3? And they're telling us that
x and y are positive. Fair enough. Statement one. x is 2 more than 3 times y. So x is equal to 3y plus 2. Let's rewrite the original
question because I think it's going to look a lot like this. If we multiply y times both
sides of the equation-- and we don't have to change the
inequality because they told us that y is positive. They're both positive. So if you multiply both sides
of the equation times y, you get x is greater than 3y. This is the same question
as this. Well, if we look at statement
number one, clearly x is greater than 3y. Because x is 2 more than 3y
based on statement number one. So statement number
one is sufficient to answer this question. x is definitely greater
than 3y. x actually equals 3y plus 2. This plus 2 tells you that
x is that much greater. 2 greater. Second statement. The ratio of 2x to 3y
is greater than 2. Let's just do some algebraic
manipulation. 3y, once again, we know
that's positive. So when we multiply both sides
of the equation by it, we don't have to switch the sign. So you get 2x is equal to 6y. Now we can divide
both sides by 2. And you get-- oh, sorry,
why did I put equals? It should be greater than. Divide both sides by 2 and you
get x is greater than 3y. So this statement and this
statement are the same thing, so 2 alone is also sufficient. So either statement alone is
sufficient to answer this or this question. 152. OK, if arc PQ-- OK, let me
see if I can draw this. So they've drawn a semicircle. So let me draw a circle. That's a circle, and I can
make it a semicircle by drawing a line from
there to there. That's about as good
as I can do. And they have it connecting
to some point. Let me see if I could do this. Looks like that. And then it looks like that. And then they actually have
drawn an altitude coming down like that. And I think I've done
all my line drawing. And then, let's see, they
labeled this P. They labeled this Q. They labeled this R. And this is 2. And they say that this distance
right here is a. This distance right here. And then they say that this
distance right here is b. Fair enough. So they say if arc PQR-- so
that's that arc right there-- is a semicircle, what is the
length of diameter PR? So they essentially want to know
what a plus b is, right? That's PR. So that's their question. So immediately-- well let's just
look at their statements. What did they give us? Statement one, they tell us
that a is equal to 4. So from knowing that a is equal
to 4, if we can figure out what b is, then we would
know what a plus b is and we would be done. Well, let's see if
we can do that. Statement two is b
is equal to 1. So if we know what b is, and
then we can figure out a from it, then we know what
a plus b is. So actually, I think if either
of these is sufficient, the other one probably is. Because if statement one is
sufficient-- if knowing a we can figure out b, then probably
knowing b we could figure out a. But let's see if that holds. So one thing that immediately
should pop out of your head, although I don't know if it
will-- and this pops out of my head, really, from going
to math competitions when I was a kid. But if you have a triangle
inscribed in a semicircle, that triangle-- and one of its
sides is a diameter, right? And that's the case
with this one. Because we know it's a
semicircle, so we know that this is the diameter--
then that triangle is a right triangle. This is a right triangle. Even the way they drew it, it
kind of looks like that. But you that's something
you should know. Even if this triangle was drawn
like that, it's always going to be a right triangle. And I'll maybe do another
YouTube video proving that. But that's just a good thing
to know for the gmat. So let's see if we could use
that information and the Pythagorean theorem to come
up with a formula that relates a and b. So if we look at just
this triangle right there, we know what? We know that a squared plus 4 is
equal to-- let's call this side c squared-- is equal
to c squared. Fair enough. If we look at this smaller
triangle right here, we know that b squared plus 4 is equal
to-- let's call this d squared, that's the hypotenuse
of the smaller triangle-- is equal to d squared. Now, if we look at the larger
triangle, which we also know is a right triangle, now c
squared and d squared are the non-hypotenuse sides, and a plus
b is a hypotenuse side. So we know that c squared plus
d squared is equal to this entire side squared, which
is a plus b squared. Well, we know what c squared
and d squared are. We can substitute that. So c squared is a squared plus
4-- and I'll switch colors just to avoid the monotnony. So c squared is a
squared plus 4. And d squared is b
squared plus 4. And all of that's equal
to a plus b squared. Let's just multiply that
out because I think we can do some canceling. So we get a squared plus
2ab plus b squared. OK. So, we could subtract a squared
from both sides. a squared, a squared. Subtract b squared
from both sides. b squared and b squared. And this simplifies
out quite nicely. Let's see. 4 plus 4 is 8, is
equal to 2ab. Divide both sides by 2, and
you get ab is equal to 4. So all of this crazy
information, that this is inscribed in the semicircle and
this is a, this is b, all of that reduces down to
ab is equal to 4. So now we know. If we know a, if a is 4,
then we know that b has to be equal to 1. And then we can figure
out what a plus b is. It's 5. So statement one alone
is sufficient. And similarly, if we know
what b is, we know that a has to be 4. And so a plus b is equal to 5. So that was actually
less hairy than I thought going into it. Next problem. And this really is
the home stretch. 153. Does the integer k have a factor
p such that 1 is less than p is less than k. OK, so they're essentially
saying, is k prime? This is another way of saying
is k-- or you could say, is k not prime? Or k is composite. I think that's the
word for it. A non-prime number. So that's all they're asking. Does k have some other factor
other than 1 and itself. So let's see, statement
number one. k is greater than 4 factorial. Well, 4 factorial is 4 times
3 times 2 times 1. So that's 12 times 2. That's 24. So this boils down to k
is greater than 24. So I can give you both prime
and non-prime numbers greater than 24. 29 is prime. 30 is not prime. So statement one, at least by
itself, is pretty useless. k can be greater than 24 and
be prime or not prime. Statement two. This looks interesting. 13 factorial-- which is a large
number, so it's hard to multiply it out-- plus 2 is less
than or equal to k, which is less than or equal to
13 factorial plus 13. So, essentially, what we need
to prove, somehow, is that every number between this
and this is not prime. And let's think about
how we can do it. Well, let's take the initial
example of 13 factorial plus 2. So 13 factorial, just so you
know, is 13 times 12, times 11-- you keep going-- times
3 times 2 times 1. That's 13 factorial. So 13 factorial is definitely
divisible by 2. So we could rewrite
13 factorial. We could write that as-- 13
factorial divided by 2 times 2 plus 2. I just rewrote 13 factorial
plus 2. And I want you to see that this
is the same thing, right? I just divided by 2 and
multiplied by 2. And you'll see in a second
why I did that. And this is a tricky problem. When I looked at this at first,
which I actually did glance at before the video,
because-- I don't know if this is something I would have gotten
right if I was under time pressure on the gmat. But when I thought about it a
little bit more it made sense. So if I divide by 2 and times 2,
I could-- and this is going to be an integer. Because we know that 13
factorial is divisible by 2. 2 is actually one
of its factors. So as long as you understand
where I got-- you can divide anything by 2 and multiply by 2
and not change it So this is the same thing as this. And then if we factor the 2 out,
this becomes 2 times 13 factorial over 2 plus 1. I just factored the 2 out. This isn't a fraction. That's just a line I did just to
not make it look too messy. So what we show is that 13
factorial plus 2 is definitely divisible by 2. Whatever that number is, it's
definitely divisible by 2. And how do I know that? Because this is just another
way of writing 13 factorial divided by 2. I'm sorry, this is another way--
let me write this down-- this is the same thing as
13 factorial plus 2. It's just another way
of writing it. I took 13 factorial divided by
2 and multiplying by 2, and then taking out the 2. And the reason why I know that
2 is divisible into this is because I know that this
is an integer. We know that 13 factorial is
divided by 2 because 2 is one of the things that
goes into it. And so this is an integer. You add 1. So this is an integer. And so you're multiplying
2 times some integer. So this has to be
divisible by 2. So 13 factorial plus 2
is divisible by 2. Now let's just extend
that generally. 13 factorial plus x. Let's prove that as long as x
is less than or equal to 13 that this is divisible by x. So we can do the same trick. You could say this is equal to
13 factorial divided by x, times x plus x. That's equal to-- you can factor
the x out-- x times 13 factorial over x plus 1. And how do we know that this
number, which is the same thing as this number, how
do we know that this is divisible by x? Well, just a little thought
experiment. We know as long as x is greater
than or equal to 1 and less than or equal to 13,
and x is an integer. As long as x is an integer
that's greater than or equal to 1 and less than or equal to
13, it is one of the factors of 13 factorial. 13 factorial is 13 times
12 times 11. It's all of those numbers
multiplied together. So we know that as long as x is
less than or equal to 13, we know that this right here,
this is an integer. If this is an integer,
then the integer plus 1 is also an integer. So 13 factorial plus x
can be rewritten as x times some integer. I'll call it I. But the bottom line is, the
reason why I wanted to do that, it shows you that every
number between 13 factorial plus 2 and 13 factorial plus
13 is actually divisible by whatever number you're adding
to the factorial. So since it's divisible by those
numbers, the numbers cannot be prime. So hopefully-- I've actually
run way over time-- but hopefully I've shown you that
that information alone, statement two alone, is
sufficient to show that k is not prime or that k does
have a factor p that's between 1 and k. Anyway, I hope didn't
confuse you. I think that was one of the
hardest problems we've done, so don't feel bad if you found
that a little daunting. See you in the next video.