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2003 AIME II problem 1

Video transcript

I got this problem here from the 2003 ami 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 to 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 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 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 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 say 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 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 we can replace C everywhere with a plus B so this expression right over here becomes a B which is just a times B times C but instead of C I'm going to write an A plus B over here a plus B and then that is equal to six times that is equal to six 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 six times a plus B plus a plus B this is the same thing as six 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 right over here is still is still a times B or a B times a plus B so a B 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 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 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 there's only so many numbers so many positive integers where you take the product you get 12 let's try them out let's try them out so let me write some columns here let's say a b c and then we care we care about their product we care about the products 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 one 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 there some there some is 26 26 times six is 156 so this one definitely worked it definitely worked for the constraints as it should because we boil down those constraints to a times B need to be equal to 12 so let's try another one two times six their sum is eight 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 & 4 3 plus 4 is 7 3 times 4 is 3 times 4 is 12 times 7 actually I should have known but a times B is always 12 so we just have to multiply 12 times 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 would have to deal with non integers you'd 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 you take their products you get you get 12 we've essentially just factored 12 so they want to they want us to find the sum of all possible values of n well these are all the possible values of N and theirs are 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