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.
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.