(Medium) Let a, b and c be nonzero integers, with 1 as their only positive common divisor, such that 1a+1b+1c=0\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0. Find the number of such triples (a,b,c)(a, b, c) with 50abc150 \geq \mid a \mid \geq \mid b \mid \geq \mid c \mid \geq 1

Note by Cai Junxiang
1 year, 2 months ago

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The answer to your question is 66 I found it by a python program do you want to see it?

Sahar Bano - 1 year, 2 months ago

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Cai Junxiang - 1 year, 2 months ago

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The answer should be an even number because if (a,b,c) is a solution, so is (-a,-b,-c) i.e. solutions come in twos. Brute force with a computer program identified these 22 solutions: -45,-36,20 -42,-7,6 -40,-24,15 -35,-14,10 -30,-6,5 -28,-21,12 -20,-5,4 -15,-10,6 -12,-4,3 -6,-3,2 -2,-2,1 2,2,-1 6,3,-2 12,4,-3 15,10,-6 20,5,-4 28,21,-12 30,6,-5 35,14,-10 40,24,-15 42,7,-6 45,36,-20 All have the pattern that a and b have the same sign and c has the opposite sign. The most common solution has the pattern (n(n+1),n+1,-n) - or the same pattern with the signs switched. There are other solutions though. I'm not sure if there is a more elegant way to solve this than brute force.

Justin Travers - 1 year, 2 months ago

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@Brilliant Mathematics, @Jake Grane has posted a comment not related to mathematics.

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