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2003 AIME II problem 1

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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.