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Reciprocal Equation #6 (Posted on 2021-02-28) Difficulty: 4 of 5
x, y, and z are each nonzero integers.

Which triplets (x,y,z) satisfy the equaiton 1/x + 2/y + 3/z = 1?

No Solution Yet Submitted by Brian Smith    
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Some Thoughts The mark of the beast (partial spoiler) | Comment 1 of 3
Not a full set of solutions -- just some obvious ones:

variations on 1/2 + 1/3 + 1/6 = 1
(2,6,18)
(2,12,6)
(3,4,18)
(3,12,6)
(6,4,9)
(6,6,6) -- The mark of the beast

variations on 1 + w - w = 1, where k is a non-zero integer
(1, 2k, -3k)
(k, 2, -3k)
(k, -2k, 3)

  Posted by Steve Herman on 2021-02-28 08:08:09
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