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

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That stands for the American Invitational[br]Mathematics Exam, and

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

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

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

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Find the sum of all possible values of N.

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

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So we have three positive integers right[br]over

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here, so let's just think about three[br]positive integers.

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Let's call them a, b and c.

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They're all positive, they're all[br]integers.

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The product N, of these 3 positive, these[br]3 positive integers.

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So a times b times c is equal to N, is 6[br]times their sum.

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This is equal to 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 three positive[br]integers is 6 times, is 6 times their sum.

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

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

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

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

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We could do, it doesn't matter, these are[br]just names and we

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haven't said one of them is larger or less[br]than the other one.

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So let's just said that a plus b is equal[br]to c.

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The one of the integers is the sum of the[br]other two, c is the sum of a plus b.

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Find the sum of all possible values of N.

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So let's just try to do a little bit of

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

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we can get some relationship or some[br]constraints on our

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

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

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So we could replace c, we can replace c[br]everywhere with a plus b, so this

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expression right over here becomes ab,[br]which is just a times b, times c, but

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instead of c, I'm gonna write an a plus b[br]over here, a plus b,

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

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a plus b plus c.

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

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And then what does this simplify to.

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

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

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just added the a's and the b's and we can[br]factor out a 2.

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This is the same thing as if you take out[br]a 2, 6 times 2 is 12 times a

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plus b, the left hand side over here is[br]still, is still a

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times b, or a b, times a plus b, so ab[br]times

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

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So this is pretty interesting here, we can[br]divide both sides by a plus b.

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We know that a plus b won't be equal to,[br]cannot be

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

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

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

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

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So all the constraints that they gave us[br]boiled down to

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this right over here, the product of a and[br]b is

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equal to 12 and there's only so many[br]numbers, so many

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positive integers where, if you take their[br]product, you get twelve.

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Let's try them out.

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Let's try them out.

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So let me try some columns here.

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

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We care about their product.

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So I'll write that over here.

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So a, b, c.

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

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

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

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

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

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Their sum is, 26, 26 times 6 is 156

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so this one definitely worked, it[br]definitely worked for the

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

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to a times b needed to be equal to 12.

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So let's try another one, 2 times 6, their[br]sum is

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8, and then if I were to take the product[br]of all

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

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Then we could try 3 and 4, 3 plus 4 is 7,[br]3

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times 4 is, 3 times 4 is 12 times 7,[br]actually I should

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

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12 times 7 is 84, 12 times 7 is 84, and[br]there

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

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you'd have to deal with the non-integers,[br]you'd have to deal with the fractions.

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

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they all have to be positive integers, so[br]that's

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it, those are all of the possible positive[br]integers,

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we take their products, you get, you get[br]12.

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You've essentially just factored 12.

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

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Well these are all the possible values of[br]n.

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

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Let's just take the sum, 6 plus 6 is 12[br]plus 4 is 16,

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1 plus 5 is 6 plus 9 is 15 plus 8 is 23,

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2 plus 1 is 3, so our

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answer is 336.