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Current time:0:00Total duration:9:51

We're on problem 145. It says, is 1/p greater than
r/r squared plus 2. That's their question. Statement number one-- I realize
I go on these tangents before even looking at the
statement; I should look at the statements first--
p is equal to r. So let's see if we can
simplify this. So if p is equal to r, then we
get 1/r-- instead of a p, right?-- is greater than
r/r squared plus 2. Let's multiply both sides
of this equation by r. Well, we don't know that r is
necessarily greater than 0. That's the problem. So if r is greater than 0--
well, it's going to change the inequality one way
or the other. So let's just assume r
is greater than 0. So if we multiply both sides of
the equation by r, then we don't have to switch the
inequality, because we're assuming r greater than 0. That's an assumption. Then you get 1 is greater
than r squared over r squared plus 2. And then if r is greater than
0, then r squared plus 2-- actually, r squared plus 2 is
always going to be greater than 0, because r squared
[INAUDIBLE] positive. Let's multiply both sides
of this equation by r squared plus 2. And you get r squared plus 2
is greater than r squared. You could subtract r squared
from both sides, and you get 2 is greater than 0,
which is true. But remember, I had to make
this assumption that r is greater than 0. If you assume that r less than
0, then all of this is going to break down. Because when you multiply both
sides by r, you have to switch the inequality to get 1 is less
than r squared over r squared plus 2. And then, eventually, you're
going to end up with r squared plus 2 is less than r squared. Or that 2 his less than
0, which is false. So this statement alone
isn't enough. We have to know whether
or not r is greater 0. Statement two-- well, there you
go; I really should look at the statements first--
r is greater 0. And just so you know, statement
two by itself is insufficient, because if you
know that r is greater than 0, you have no idea what p is. So you can't answer it yet. But both statements combined
are sufficient to answer this question. Next question. 146. Is n an integer? Statement one says n squared
is an integer. Well, that doesn't help us. I mean, what if n was equal
to the square root of 2? And then n squared
would be equal to 2, which is an integer. But this is clearly
not an integer. But on the other hand, n could
be equal to 2, in which case n squared would be an integer. So whether or not n squared is
an integer, both of these are cases where n squared are
integers, but one ends up where n is an integer, one is
where n isn't an integer. So this, by itself, is not
enough to tell me whether n is an integer. Statement two tells
us, the square root of n is an integer. OK. So that essentially tells us
that n is equal to some integer squared, right? You could take the square
root of both sides. You'd get the square root of
n is equal to some integer. So you take any integer, you
square it, you're going to get an integer. So statement two alone
is sufficient to answer this question. That was a strangely
easy question. They they were getting a
little bit hairy and confusing, but that was a nice
little rest question, I think. 147. If n is a positive integer--
so n is a positive integer. Well, n is greater than 0. It's an integer. Well, I didn't write
that down. Is n to the third minus
n divisible by 4? Fascinating. Let's see. So my initial reaction is to see
if I can simplify this as a product of simpler
expressions. I look already at statement
number one. Let me look at statement
number one. Statement number one says, n is
equal to 2k plus 1, where k is an integer. I don't feel like taking 2k
plus 1 and cubing it. That's not an easy
thing to do. It'll take some time. So my intuition is that maybe
we should simplify this a little bit. So if we factor out an n, that's
equal to n times n squared, minus 1. And that's equal to n times--
what's n squared minus 1? That's n plus 1, times
n minus 1. And now this is something
that's much easier to substitute this into. So let's do that. Let's substitute n equals
2k plus 1 into this. So if n is equal to 2k plus 1,
you get 2k plus 1, times 2k plus 1, plus 1. So that's 2k plus 2. That's this one. And then you have 2k
plus 1, minus 1. So that's just 2k. And then what does
that simplify to? You get 2k plus 1. Remember, in the back my mind, I
want a 4 to show up, because I want this thing to
be divisible by 4. 2k plus 1, times 4k
squared plus 4k. Interesting. So now I can factor out a 4. So it's 2k plus 1, times k
squared plus k, times 4. And we know that this is an
integer, because they told us that k is an integer. We know that this
is an integer, because k is an integer. And we know 4 is an integer. So we've essentially kind of
factored n cubed minus n. And 4 is one of it's factors. So it's definitely
divisible by 4. I mean, we can divide
it by 4 right now. If we divide it by 4, we'd be
left with this, which is clearly an integer, because
k is an integer. So n to the third minus n is
divisible by 4 if we can assume that n is equal
to 2k plus 1, where k is some integer. Now, what do they tell
us in statement two? Statement two tells us,
n squared plus n is divisible by 6. Well, I don't see how that's
helpful at all. n squared plus n is
divisible by 6. You can't even relate
that to that. I mean, n squared--
I could try to do something fancy here. No. This is so different than this.
n squared plus n is so different from n to the third. I could factor out an n. I could say n times n plus
1 is divisible by 6. Well, I guess that's
interesting. That tells us that this part
is divisible by 6. n times n plus 1 is
divisible by 6. Do they tell us that n is
a positive integer? Yeah, they do tell us that
n is a positive integer. So this could be 2 and 3. If this is 2 and 3, then
this would be 1. This could be 2 and 3. In fact-- well, this could be
2 and 3, but it could also be-- well, actually, that's the
only numbers it could be. It could be 2 and 3, in which
case this is 1, in which case this whole thing-- if you say
this, then n could be 2. n is equal to 2. n plus 1 is equal to 3. And then n minus 1 would
be equal to 1. And let me see. Is that the only case? I'm looking at the answer right
now, just because I want to make sure. They say that this
isn't sufficient. But, once again, kind of like
problem 142, actually-- because if you say that n is a
positive integer and that n times n plus 1 is divisible by
6, I can't think of any other numbers where you multiply one
number times the next number, and their integers, where
you get a multiple of 6. No, I take that back. It could be 3 and 4. There could be 3 and 4. It could be-- yeah, there's
actually a bunch of them. This could be 3, this could be
4, because it's a multiple 6, and then this would be 2. So that's right. This is not enough
information. So that's good. Because by this information, n
could be 2, n plus 1 could be 3, in which case, this number
would be actually 6. But that still doesn't help
us, because 6 isn't divisible by 4. But then I could come up with
the situation where this number right here, where n is 3,
n plus 1 is 4, and n minus 1 is 2, where this does
become divisible by 4. So this doesn't give us
enough information. Not useful. And I'm out of time. I'll see you in the
next video.