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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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Solution re: computer solution Comment 3 of 3 |
(In reply to computer solution by Charlie)

Well, I noticed my program missed, for example, 2,6,18, mentioned by Steve Herman. I found the bug: I had used gcd of the denominators for multiplying the numerators, supposedly to assure not getting rounding problems in 1/2 + 2/6 + 3/18, but I should have used the lcm to avoid extended fractions voiding the equality condition.  I redid the solutions for all x, y, z positive:


     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
 
    

Edited on February 28, 2021, 2:30 pm
  Posted by Charlie on 2021-02-28 10:07:32

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