2003 AIME II Problem 1

2003 AIME II Problem 1: The product N

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 six 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 3 positive, these
3 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, is 6 times their sum.

So, this is equal to six times the sum of
those integers, a plus b plus c.

And one of the integers is the sum of the
other two.

One, 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.

We could do, 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 said that a plus b is equal
to c.

The 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 of what we
have here and maybe,

we can get some relationship or some
constraints on our

numbers and then we can kinda go through
all the possibilities.

So let's see, we know that a plus b is
equal to c.

So we could replace c, 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 gonna write an a plus b
over here, a plus b,

and then that is equal to 6 times, that is
equal to 6 times a plus b,

a plus b plus c.

And so, once again I'll replace the c with
an a plus b.

And then what does this simplify to.

So on the right hand side, we have 6 time
a plus b plus a plus b.

This is the same thing as 6 times 2a plus
2b, 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 over here is
still, is still a

times b, or a b, 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 won't be equal to,
cannot be

equal to zero since all of these numbers
have to be positive numbers.

So if we divide both sides, and the reason
why I say that is you,

if you divide, if it was zero, dividing by
zero would give you an undefined answer.

So if we divide both sides by a plus b, we
get a times b is equal to twelve.

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, if you take their
product, you get twelve.

Let's try them out.

Let's try them out.

So let me try some columns here.

Let's say a, b, c, and then we care, we
care about their product.

We care about their product.

So I'll write that over here.

So a, b, c.

So if a is 1, if a is 1, b is going to be
12, c is the sum of

those two so c is going to be 13, 12, 1
times

12 times 13, 12 times 12 is 144, plus
another 12 is going to be 156.

And just out of, 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 needed 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, 96.

Then we could try 3 and 4, 3 plus 4 is 7,
3

times 4 is, 3 times 4 is 12 times 7,
actually I should

have known, a times b is always 12 so we
just have to multiply 12 this last column.

12 times 7 is 84, 12 times 7 is 84, and
there

aren't any others, you can't go, you
definitely can't go above 12, because then

you'd have to deal with the non-integers,
you'd have to deal with the 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,

we take their products, you get, you get
12.

You've essentially just factored 12.

So, they want us, they want us to find the
sum of all possible values of N.

Well these are all the possible values of
n.

N is the product of those integers, so
let's just take.

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.