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# 2003 AIME II problem 1

Video transcript

I got this problem here
from the 2003 AIME exam. That stands for the American
Invitational Mathematics Exam. And this was actually the
first problem in the exam. The product N of three positive
integers is 6 times their sum, and one of the integers is
the sum of the other two. Find the sum of all
possible values of N. So we have to deal with
three positive integers. So we have three positive
integers right over here. So let's just think about
three positive integers. Let's call them a, b, and c. They're all positive. They're all integers. The product N of these
three positive integers-- so a times b times c is equal
to N-- is 6 times their sum. This is equal to
6 times the sum. Let me do this in another color. So this is their product. So the product N of
three positive integers is 6 times their sum. So this is equal to 6
times the sum of those integers a plus b plus c. And one of the integers is
the sum of the other two. Well, let's just pick c
to be the sum of a and b. It doesn't matter. These are just names, and
we haven't said one of them is larger or less
than the other one. So let's just say
that a plus b is equal to c, that
one of the integers is the sum of the other two.
c is the sum of a plus b. Find the sum of all
possible values of N. So let's just try to do a
little bit of manipulation of the information we
have here, and maybe we can get some relationship
or some constraints on our numbers, and
then we can kind of go through all of
the possibilities. So let's see, we know that
a plus b is equal to c. So we can replace c
everywhere with a plus b. So this expression
right over here becomes ab, which is
just a times b, times c. But instead of c, I'm going to
write an a plus b over here. And then that is equal to
6 times a plus b plus c. And so once again, I'll replace
with the c with an a plus b, and then what does
this simplify to? So on the right-hand side,
we have 6 times a plus b plus a plus b. This is the same thing
as 6 times 2a plus 2b, just added the a's and the b's. And we can factor out a 2. This is the same thing as if
you take out a 2, 6 times 2 is 12 times a plus b. The left-hand side
right over here is still a times b
or ab times a plus b. So ab times a plus b has got to
be equal to 12 times a plus b. So this is pretty
interesting here. We can divide both
sides by a plus b. We know that a plus b cannot be
equal to 0 since all of these numbers have to be
positive numbers. And the reason why I say that
is if it was 0, dividing by 0 would give you an
undefined answer. So if we divide both
sides by a plus b, we get a times b is equal to 12. So all the constraints
that they gave us boiled down to this
right over here. The product of a and
b is equal to 12. And there's only
so many numbers, so many positive integers where
you if you take the product, you get 12. Let's try them out. So let me write
some columns here. Let's say a, b, c. And then we care
about their product, so I'll write that
over here, so abc. So if a is 1, b
is going to be 12. c is the sum of those two, so
c is going to be 13, 1 times 12 times 13. 12 times 12 is 144 plus
another 12 is going to be 156. And just for fun, you
can verify that this is going to be equal
to 6 times their sum. Their sum is 26,
26 times 6 is 156. So this one definitely worked. It definitely worked
for the constraints. And it should because we
boiled down those constraints to a times b need
to be equal to 12. So let's try another one. 2 times 6, their sum is 8. And then if I were to take
the product of all of these, you get 2 times 6
is 12 times 8 is 96. And then we could try 3 and 4. 3 plus 4 is 7. 3 times 4 is 12 times 7. Actually, I should have known
the a times b is always 12, so you just have to multiply
12 times this last column. 12 times 7 is 84. And there aren't any others. You definitely can't go
above 12 because then you would have to deal
with non-integers. You would have to
deal with fractions. You can't do the negative
versions of these because they all have
to be positive integers. So that's it. Those are all of the
possible positive integers where if you take their
products you get 12. We've essentially
just factored 12. So they want us to find the sum
of all possible values of N. Well, these are all
the possible values of N. N was the product
of those integers. So let's just take the sum. 6 plus 6 is 12 plus
4 is 16, 1 plus 5 is 6 plus 9 is 15 plus
8 is 23, 2 plus 1 is 3. So our answer is 336.