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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 value of $N$.

Based the given, I think the sum would be 2 because $N$ itself is 2. However I don't know if this is correct.

$ N=ab(a+b)=12(a+b)$

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$$6(a+b+c) = abc$$

$$a+b=c$$

$$6(a+b+a+b) = ab(a+b)$$

$$12(a+b)=ab(a+b)$$

$$ab = 12$$

$$a = 3, b = 4, c = 7$$

$$a = 2, b = 6, c = 8$$

$$a = 1, b = 12, c = 13$$

$$N = 84, 96, 156$$

Not sure if this answers the question but the sum of all possible values of $N$ is therefore $341$.

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  • $\begingroup$ Thanks,This is the answer I am looking for. I saw the careless mistake I made. $\endgroup$ – most venerable sir May 6 '14 at 23:24
  • $\begingroup$ @shevliaskovic - thanks for the edit, I was using my phone and it's hard enough to type without the formatting! $\endgroup$ – mjsqu May 7 '14 at 9:00
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Hint: Let $x,y,z$ be the three positive integers. Then, $$xyz=6(x+y+z)\\ x+y=z\\ \implies xyz=6(z+z)=12z\\ \implies xy=12$$

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