Main content

## Math for fun and glory

### Course: Math for fun and glory > Unit 4

Lesson 1: 2003 AIME- 2003 AIME II problem 1
- 2003 AIME II problem 3
- Sum of factors of 27000
- Sum of factors 2
- 2003 AIME II problem 4 (part 1)
- 2003 AIME II problem 4 (part 2)
- 2003 AIME II problem 5
- 2003 AIME II problem 5 minor correction
- Area circumradius formula proof
- 2003 AIME II problem 6
- 2003 AIME II problem 7
- 2003 AIME II problem 8
- Sum of polynomial roots (proof)
- Sum of squares of polynomial roots
- 2003 AIME II problem 9
- 2003 AIME II problem 10
- Trig challenge problem: area of a triangle
- 2003 AIME II problem 12
- 2003 AIME II problem 13
- Trig challenge problem: area of a hexagon
- 2003 AIME II problem 15 (part 1)
- 2003 AIME II problem 15 (part 2)
- 2003 AIME II problem 15 (part 3)

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# 2003 AIME II problem 10

perfect squares integers square root. Created by Sal Khan.

## Want to join the conversation?

- What is AIME I live in the UK so I am not familiar with all the American exam jargon. If someone could post a brief overview I would be very grateful.(4 votes)
- Here is all the info that you perhaps need:

http://en.wikipedia.org/wiki/American_Invitational_Mathematics_Examination(8 votes)

- he says that 4*125 is 900 im not sure about that(5 votes)
- Yes it's an error but it doesn't relate to solving the question, since 225 is still odd and the constraint was to have two even numbers.(1 vote)

- Hey Sal, just curious, did u really work out all this on the spot ? This was a veryy good problem. Any suggestions on how can we build ourselves and work out such problems while giving the actual exam? Thanx :)(1 vote)
- Not on topic, but is there any public class for me to join?(1 vote)
- At5:01, we realize that (b)(b + 60) = b^2 + 60b has to be a perfect square. To find the values of b that make that expression a perfect square, Sal completes the square and then rearranges to obtain a difference of squares. Is there any other approach? My first attempt was to rewrite b^2 + 60b = n^2 as b^2 + 60b - n^2 = 0 and then use the discriminant to find the correct values of b (the discriminant must also be a perfect square), but the numbers involved were very large and this approach took a lot of time. In this case my approach wasn't very useful, so my question is how would I know when it is easiest to solve using a difference of squares and when it is easiest to solve using the discriminant (or any other simpler method)?(1 vote)

## Video transcript

Two positive integers
differ by 60. The sum of their square
roots is the square root of an integer that is
not a perfect square. What is the maximum possible
sum of the two integers? So the second sentence
is a little confusing, but let's take a step-by-step. So two positive
integers differ by 60. So let's say they're a and b. So we have a and b. They're both greater than 0. They're also both integers. And they differ by 60. So a minus b is equal to 60. Or we could add b to
both sides, and we could get a is
equal to b plus 60. They're 60 apart. Fair enough. Now this second sentence, the
sum of their square roots-- the square root of a
plus square root of b-- the sum of the square
roots is the square root of an integer that is
not a perfect square. So it's equal to the square
root of some integer. So c is an integer, but
it's not a perfect square. So square root of c
is not an integer. So that's the second part. What is the maximum possible
sum of the two integers? So if we could maximize b
and find an a that fits it, or an a and a b that
meet these constraints we could get them
as big as possible, then the sum of a and b
will also be maximized. So let's think about
that in a second. But just to simplify-- this is
actually the really confusing part because c is an
integer and square root of c is not an integer. And to make it a
little less confusing, let's just square both sides
of this equation over here. So square root of a
squared is-- so we're just going to square both sides. So this is going
to be square root of a times square
root of a-- which is a-- plus 2 times the
square root of a times the square root of b-- which is
2 times the square root of ab-- plus the square root of
b squared-- which is just going to be b-- is equal to c. Now, we know that
a is an integer. We know that b is an integer. And we know that
c is an integer. We know that these are integers. So if I have an
integer plus something plus an integer equaling
another integer, this thing in the middle
has to be an integer. And actually, I'll show you
right now that not only does this thing have to be an
integer, square root of ab has to be an integer. And you might be saying, wait,
wait, wait, wait a second. If this thing just
has to be an integer, can't I just say-- let me do
it over here on the side-- can't I just say that 2
times the square root of ab has to be equal to some integer. You could divide
both sides by 2. And you could say, well,
maybe the square root of ab is equal to k/2, where
this is not an integer. I claim that this thing
does have to be an integer. But you say, hey look, just
straight from the math, this could be an integer. But maybe divide
both sides by 2. And this doesn't have
to be an integer. And if this is not
an integer, that means that k is odd,
because otherwise it would be divisible by 2. Now, if we square
both sides of this, we would get ab is equal
to k squared over 4. Well, this is going to
be an odd times an odd. It's going to be an odd squared. So this is still
going to be odd, which is not divisible by 4. So this thing on
the right hand side will still not be an integer. But remember, a
and b are integers. So a times b, this
is an integer. So if a and b are
integers, there's no way that the square
root of its product can be something divided by
2, where this doesn't reduce any more, where this
doesn't simplify, or some odd number divided by 2. So we just get a
contradiction here. Or another way to say it
is the square root of ab-- because this whole thing has
to be an integer and both a and b are integers,
the square root of ab has to be an integer. But if the square root of
ab has to be an integer, then that means that ab
has to be a perfect square. So then I'll start over here
so I don't run out of space. Actually, let me do
it right under that because it's a straight result. So that means that a times
b is a perfect square. So let's see how we can
deal with this a little bit. We can actually do a
substitution for a. a is equal to b plus 60. So this means that
b plus 60 times b is equal to a
perfect square, so it could be equal to some number n
squared, where n is an integer. And let's see if we can play
around with this a little bit, put some constraints
on what we're doing. So if you multiply
this out, you get b squared plus 60b-- I'll
leave some space over here-- is equal to n squared. And let's see, we could
complete the square over here. And maybe we could
subtract this over and we'll have a
difference of squares. So we want to complete
the square here. You take half of this. So half of 60, half
of the coefficient-- that's 30-- and square it. So this would be
plus 900 minus 900. We haven't changed it. We're adding and
subtracting the same thing. But what that does for us
is that this part over here becomes b plus 30 squared,
and then we have the minus 900 is equal to n squared. And then we could subtract
n squared from both sides and add 900 to both sides. And we get b plus 30--
and you really just have to play around
with this thing and see what
constraints you can get. b plus 30 squared minus n
squared is equal to 900. And this is just a
difference of squares. So this can be factored. This is b plus 30. This is a plus b. Or, I shouldn't use a and b. But it's the square root of this
plus the square root of this, so plus n. Times the square root of this
minus the square root of this. So, b plus 30 minus n. So we can factor
it just like that. And that's about as good of
a constraint as we can get. Remember, we want to
be able to maximize b. If we're maximizing b,
a is only 60 above b. So we're also going
to be maximizing a. And so we'll also be
maximizing the sum. So we want b to be
as high as possible. So b plus 30 plus n is going
to be some factor of 900. And b plus 30 minus n will
be the other factor of 900. And we could actually do a
little bit more of an insight here. We have two factors. Their product is an even number. So it's an even product. And if we were to take the
sum of these two factors, if you were to take
b plus 30 plus n and add it to b
plus 30 minus n-- so we're just taking the
sum of these two factors-- we get 2b plus 60. And these cancel out. And so this is also even. So if I take two numbers,
and their product is even, and their sum is
also even, that means the two numbers have to be even. So both this guy and this guy
over here have to be even. Both of them have to be even. And at this point, I think we've
used up all the information in the problem. We really just have to try
out factors of 900 that meet these constraints
where both are even. And let's try with the most
extreme things in an attempt to maximize our b. So let's think about
the factors of 900. You could do 1 and
900, but remember, both factors have to be even. 1 is not even. We could try 2 and 450, where
this will be the higher factor. This will be the lower
factor because we're subtracting a number here. So let's try that out. Actually, let me
do it over here. Actually, I'll do it over here. Maybe this is 450 and this is 2. So we have b plus 30
plus n is equal to 450, and that b plus 30
minus n is equal to 2. And now we can solve for b. We can add these two equations. You get 2b plus 60-- these
cancel out-- is equal to 452. Subtract 60 from both sides. 2b is equal to 392,
or b is equal to 196. So let's see how this works out. If b is equal to 196, then
a is just 60 plus that. So a is equal to 256. And let's see what it does
to this equation over here. Square root of a is 16. So we would have the
square root of 256 plus the square root of 196. Well, that's equal to 16 plus
14, which is equal to 30. So it looks like it might work. But remember, this thing over
here, the square root of c, is not a perfect square. This value right over here. That's the square root of c. Or I should say the square
root of c is not an integer. It cannot be an integer. If it was, then c which
in this case would be 900, would be a perfect square. But they told us in the problem
c is not a perfect square. So this one actually does not
work because this would give us a c that is a perfect square. It gives us 900. Square root of c is 30. So we have to cross
these out too, although it was a very
tantalizing solution since it gave us
nice clean numbers. So let's keep trying
factors of 900. So remember, you
could try 3 and 300, but we know that they
both have to be even. So we could just reject
this because 3 is not even. So maybe it's 4 and-- 4
goes into 900 125 times. You might be tempted
to try these out. But 125 is not even. So we could keep going. So then we could go to 5. Well, once again 5 is not
even so that won't work. We could try 6. 6 goes into 900-- it
goes into 90 15 times. So it goes into 900 150 times. They're both even numbers. So maybe, just maybe
these will work. So let's try it out again. So let me scroll
down a little bit. So you could have
b plus 30 plus n-- that'll be the larger
number-- is equal to 150 because we're
adding the n there. And then you have b plus 30
minus n would be equal to 6. You add both equations, you
get 2b plus 60 is equal to 156. Or subtracting 60
from both sides 2b is equal to-- this
is equal to 96. Or b is equal to 48. So here's our next candidate. Our next candidate
is b is equal to 48. Then a would be
60 more than this. a would be equal to
108, 60 more than that. If I were to take the
square root of 108 plus the square root of
48, what does that give me? 108 is 36 times 3. So this is the square root of
36 times the square root of 3, or 6 square roots of 3. Plus-- this is 16 times 3. So it's going to be the
square root of-- well 48 is 16 times 3. So it's the square root of 16
times the square root of 3. So it's 4 square roots of 3. So when you add
these two together, you get 10 square roots of 3. So this fits our bill
for the square root of c. So over here we
have the square root of c is equal to 10 roots of 3. It's definitely not an integer. So it makes that. And if you square
it, you would get c as being equal to 100 times
3, which is equal to 300. And it's an integer. So it works. So our best shot, our best
values for a and b, b is 48, a is 108. If we take the sum, 108 plus
48, we get 16, 5, and 156. And you can see when you
keep trying-- obviously we didn't have the best result
when we had b being 196. But as we get our factors
closer and closer together, the b is going to go
lower and lower and lower. And we're obviously
going to have a less of a maximum
value for a plus b. So as we got from the most
extreme factors of 900, we keep going until
something met. This met our constraints. If we kept going, if
we tried eight and-- I don't know if that would
even work as a factor. But if we just tried and kept
looking at the other factors, we actually would
have gotten numbers that are closer to each other. And that would have led
to lower values for our b. So this right here
is our maximum b-- or the b that will
maximize our result. And the sum is 156.