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Positive Integer Solutions

Find all positive integer solutions \((x,y,z)\) that satisfy: \[\frac1x +\frac2y+\frac3z=1\]

Note by Idham Muqoddas
4 years, 5 months ago

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A tedious solution would be to use the fact that \(\min \{x,y,z\} \le 6\) and solve 12 different equations in two variables of the form \[\frac{a}{u} + \frac{b}{v} = \frac{p}{q} \] (which I believe should present no problems). I presume you have an elegant way of reducing the tedium?

Peiyush Jain - 4 years, 5 months ago

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Any complete solution would still have to find all of the following 32 solutions:

(2,5,30), (2,6,18), (2,7,14), (2,8,12), (2,10,10), (2,12,9), (2,16,8), (2,28,7), (3,4,18), (3,6,9), (3,12,6), (3,30,5), (4,3,36), (4,4,12), (4,8,6), (5,4,10), (5,10,5), (5,40,4), (6,3,18), (6,4,9), (6,6,6), (6,24,4), (8,4,8), (8,16,4), (10,5,6), (12,3,12), (12,12,4), (14,4,7), (15,6,5), (20,10,4), (30,3,10), (36,9,4)

So, there probably isn't a less tedious way to do it than the method you described.

Jimmy Kariznov - 4 years, 5 months ago

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Since 0 in denominator of a fraction makes it to infinity the equation does not hold any sense. And also fraction is not an integer. There this equation has no integer solution!!!!! According to my view!

Subhrodipto Basu Choudhury - 4 years, 5 months ago

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You don't have to let any one of them be \(0\), in fact, they are all positive. Also, several fractions can have a sum that is an integer, for example,

\[ \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1. \]

Tim Vermeulen - 4 years, 5 months ago

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Of course there exists, for example, x = y = z = 6.

Sawarnik Kaushal - 4 years, 5 months ago

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