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I got this problem here[br]from the 2003 AIME exam.

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That stands for the American[br]Invitational Mathematics Exam.

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And this was actually the[br]first problem in the exam.

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The product N of three positive[br]integers is 6 times their sum,

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and one of the integers is[br]the sum of the other two.

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Find the sum of all[br]possible values of N.

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So we have to deal with[br]three positive integers.

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So we have three positive[br]integers right over here.

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So let's just think about[br]three positive integers.

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Let's call them a, b, and c.

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They're all positive.

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They're all integers.

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The product N of these[br]three positive integers--

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so a times b times c is equal[br]to N-- is 6 times their sum.

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This is equal to[br]6 times the sum.

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Let me do this in another color.

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So this is their product.

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So the product N of[br]three positive integers

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is 6 times their sum.

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So this is equal to 6[br]times the sum of those

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integers a plus b plus c.

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And one of the integers is[br]the sum of the other two.

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Well, let's just pick c[br]to be the sum of a and b.

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It doesn't matter.

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These are just names, and[br]we haven't said one of them

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is larger or less[br]than the other one.

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So let's just say[br]that a plus b is

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equal to c, that[br]one of the integers

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is the sum of the other two.[br]c is the sum of a plus b.

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Find the sum of all[br]possible values of N.

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So let's just try to do a[br]little bit of manipulation

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of the information we[br]have here, and maybe we

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can get some relationship[br]or some constraints

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on our numbers, and[br]then we can kind of

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go through all of[br]the possibilities.

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

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So we can replace c[br]everywhere with a plus b.

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So this expression[br]right over here

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becomes ab, which is[br]just a times b, times c.

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But instead of c, I'm going to[br]write an a plus b over here.

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And then that is equal to[br]6 times a plus b plus c.

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

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and then what does[br]this simplify to?

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So on the right-hand side,[br]we have 6 times a plus b

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plus a plus b.

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This is the same thing[br]as 6 times 2a plus 2b,

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just added the a's and the b's.

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And we can factor out a 2.

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This is the same thing as if[br]you take out a 2, 6 times 2

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is 12 times a plus b.

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The left-hand side[br]right over here

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is still a times b[br]or ab times a plus b.

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So ab times a plus b has got to[br]be equal to 12 times a plus b.

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So this is pretty[br]interesting here.

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We can divide both[br]sides by a plus b.

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We know that a plus b cannot be[br]equal to 0 since all of these

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numbers have to be[br]positive numbers.

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And the reason why I say that[br]is if it was 0, dividing by 0

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would give you an[br]undefined answer.

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So if we divide both[br]sides by a plus b,

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we get a times b is equal to 12.

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So all the constraints[br]that they gave us

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boiled down to this[br]right over here.

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The product of a and[br]b is equal to 12.

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And there's only[br]so many numbers, so

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many positive integers where[br]you if you take the product,

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you get 12.

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Let's try them out.

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So let me write[br]some columns here.

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Let's say a, b, c.

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And then we care[br]about their product,

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so I'll write that[br]over here, so abc.

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So if a is 1, b[br]is going to be 12.

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c is the sum of those two, so[br]c is going to be 13, 1 times

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12 times 13.

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12 times 12 is 144 plus[br]another 12 is going to be 156.

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And just for fun, you[br]can verify that this

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is going to be equal[br]to 6 times their sum.

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Their sum is 26,[br]26 times 6 is 156.

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So this one definitely worked.

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It definitely worked[br]for the constraints.

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And it should because we[br]boiled down those constraints

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to a times b need[br]to be equal to 12.

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So let's try another one.

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2 times 6, their sum is 8.

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And then if I were to take[br]the product of all of these,

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you get 2 times 6[br]is 12 times 8 is 96.

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And then we could try 3 and 4.

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3 plus 4 is 7.

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3 times 4 is 12 times 7.

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Actually, I should have known[br]the a times b is always 12,

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so you just have to multiply[br]12 times this last column.

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12 times 7 is 84.

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And there aren't any others.

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You definitely can't go[br]above 12 because then you

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would have to deal[br]with non-integers.

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You would have to[br]deal with fractions.

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You can't do the negative[br]versions of these

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because they all have[br]to be positive integers.

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So that's it.

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Those are all of the[br]possible positive integers

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where if you take their[br]products you get 12.

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We've essentially[br]just factored 12.

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So they want us to find the sum[br]of all possible values of N.

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Well, these are all[br]the possible values

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of N. N was the product[br]of those integers.

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So let's just take the sum.

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6 plus 6 is 12 plus[br]4 is 16, 1 plus 5

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is 6 plus 9 is 15 plus[br]8 is 23, 2 plus 1 is 3.

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So our answer is 336.