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Current time:0:00Total duration:10:13

We're on problem 148. What is the tens digit of
positive integer x. We want to know the
tens digit. Statement one tells us x
divided by 100 has a remainder of 30. Well, then we know the tens
digit is 3, right? Because, let's say 30 divided by
100 has remainder of 30 and its tens digit is 3. 130 divided by 100 has a
remainder of 30 and its remainder is 3. And I can make an argument
that any number, when you divide it by 100 that has a
remainder of 30, its tens digit is going to be 3. In fact, that number is
going to be 3, 0. And its going to have a
bunch of other digits here, there, there. And if you just divide this by
100, you're going to be left over with 30. So statement one is sufficient
to figure out what the tens digit of the number is. Statement two. So we could write x divided
by 110 has a remainder equal to 30. Well, this is a little
different case. So, definitely, when 30 is
divided by 110 the remainder is 30, because it goes
into it 0 times. So the remainder is 30. And then the tens digit is 3. So let's see, I'm going to try
to find other numbers when they're divided by 110 its
remainder is 30 but its tens digit is something else. So let's see. 140. If I divide that by 110, I'm
left with a remainder of 30. But its tens digit is what? Its tens digit is 4, right? I shouldn't draw
an arrow here. If I did-- let's see, I
go up one more-- 250. 250, you divide it by 110,
its remainder is 30. But its tens digit is 5. So this statement gives
me no information. Saying that the remainder
is 30 still gives me no information. The tens digit can be any of
these numbers, or a bunch more if I just kept going. So statement one is sufficient
to answer this question. Question 149. If x, y, and z are positive
integers, is x minus y odd? In order for this to happen,
one of these have to be odd and one have to be even. And you can just think
about that. If you take the difference,
or really the sum, of two numbers, the only way that
that difference or sum is going to be odd is if one
is odd and one is even. So let's think about that. Let's look at the statements. Just going to see if they
give us any information. They say that x is equal
to z squared. Well, that gives me no
information about y. And it gives me, actually, very
little information about x just yet. I mean, it tells me that x is
a perfect square, but a perfect square could be
odd or even, right? It could be 16, it could be 9. 16's an even perfect square,
9 is an odd one. So this doesn't give me much
information by itself. Let's see what statement
two tells us. Statement two says that y is
equal to z minus 1 squared. So this statement, by itself, is
kind of like statement one. It just tells me that y is
a perfect square of some integer, right? Because z is an integer, so
z minus 1 is an integer. So it just tells you y
is a perfect square. And a perfect square could
be even or odd. But if we take both of these
together, then something interesting happens. For example, if we assume that z
is odd, then its square will also be odd, right? 3 squared is 9, 7
squared is 49. So if z is odd, then x is odd. And then z minus 1 would be even
and then y would be even. So one would be odd,
one would be even. And you could do
the other way. You could say if z is
even, x is even. And if z is even, then
z minus one is odd. And y is odd. And I can prove it to
you mathematically. Let's write this and
substitute for z. So x minus y becomes z squared
minus z minus 1 squared. And so this becomes z squared
minus z squared minus 2z plus 1. And that equals z
squared minus z squared plus 2z, right? Distribute the negative sign. Plus 2z minus 1. The z squares cancel out and
we're left with 2z minus 1. If we use both statements,
x minus y simplifies to 2z minus 1. They told us that z is
a positive integer. So this-- this part of
the statement right here-- has to be even. It's a multiple of 2. So this is even. And if you subtract 1 from an
even number, this whole expression has to be odd. So both statements together are
sufficient to say that x minus y is odd. Next problem. This one looks hairy. 150. Henry purchased three
items during a sale. He received a 20% discount off
the regular price off the most expensive item and a 10%
discount off the regular price of the other two. Was the total amount of the
three discounts greater than 15% of the sum of the
regular price? OK. So what was the total amount
of the three discounts? It was 20% times-- let's just
call it item one-- 20% times item one plus 10% times
item too, plus 10% times item three. And I'm assuming this is
the most expensive. Let's just say this is the
second most, I guess. This is third most expensive. Maybe they're the same
price, I don't know. This is the total discount. This isn't what he paid. This is the discount
on item one. The discount on item two. I'm not saying what they
actually paid for item two. So what the question is, was
a total amount of the three discounts, that's this number,
greater than 15% of the sum of the regular prices. So they're saying, was that
greater than 15% of I1, plus I2, plus I3. And I think we can
simplify this. Because if we distribute this
right-hand side, you get 0.15 times I1 plus 0.15 times
I2 plus 0.15 times I3. And let's see, if we subtract
out 0.15 I1 from both sides, you get point-- let's see, 0.2
minus 0.15 is 0.05 I1. And I want to keep everything
positive, so let me subtract these from the right-hand
side. So I'm going to subtract 0.1
I2 from-- I'm going to subtract these from both
sides of the equation. So that's going to be greater
than 0.15 I2 minus 0.1 I2. So that's 0.05 I2. Plus, and now to do the
same thing for I3. 0.15, 0.1. So I'm going to subtract
0.1 from 0.15. So 0.05 I3. All I did, going from this to
this, is I just distributed the 0.15 and I subtracted
and added to simplify it a little bit. And, actually, this is
interesting, too, because I just have these 0.05's
everywhere. That's a positive number. And I was able to do that
because I just added and subtracted from both sides. But if I multiply or divide by
a positive number, then I don't have to change the
inequality, and you could say, let's just divide both sides
of this equation by 0.05. Or the equivalent is to multiply
both sides by 20. But then we're left with I1 is
greater than I2 plus I3. And I think this simplification,
without having read the statements,
was worth it. Because we went from something
very convoluted in a very convoluted problem statement
to something very simple. So, essentially, they're asking
us was the price of I1, was the price of item one,
greater than the price of item 2 plus item 3. If we can answer this question,
we can answer the harder question. So statement number one. The regular price of the most
expensive item was $50. I1 was equal to $50. And the regular price
for the next most expensive item was $20. So I2 was equal to $20. So now the question boils down
to, was $50 greater than $20 plus the third most
expensive item. Well, I don't know, depends. Well, actually, this answers
our question, right? Because I was about to say, well
the third most expensive item, maybe it's $30. But by definition, we know
that it's not $30. Why? Because it was the third
most expensive item. The second most expensive
item is $20. So this thing has to be less
than $20 if we are to consider it the third most
expensive item. So if this thing is less than
$20, then the right-hand side of the equation is definitely--
it's going to actually be less than $40. So this is definitely going to
be less than $50, so this is going to be true. So it turns out that statement
one by itself is sufficient. Because you just have to realize
that I3 has to be less than $20 I2 is equal to $20. Statement two. The regular price of the least
expensive item was $15. So let's see. That statement by itself, we get
I1 is greater than I2 plus $15, where this is the
least expensive item. So just looking at this, we know
this is going to be $15, this is going to be more than
$15-- yeah, this is hard. You can't say anything about
this, because maybe I1 is $17. Maybe I2 is $16 plus $15. And in this case, it would
not be the case. Or maybe I1 is $170 and I2 is
$16 plus $15, in which case it would be the case. So statement two, by itself,
isn't sufficient. So the answer to this is a. Statement one alone
is sufficient to answer this question. See you in the next video.