What are all the ordered triples of positive integers (x,y,z) with x<y<z, such that their product is three times their sum?
Rewrite the equation as 1/xy + 1/yz + 1/xz = 1/3
Since x < y < z, 1/x > 1/y > 1/z and 1/x^2 > each term on LHS. Then 3/x^2 > 1/3 implying x^2 < 9, and thus x = 1 or 2.
Case 1: x = 1
The equation becomes yz - 3y - 3z = 3 which factors (y - 3)(z - 3) = 12.
With y < z, the possible factor pairs are (1, 12), (2, 6), and (3, 4).
(1, 12): y = 4, z = 15
(2, 6): y = 5, z = 9
(3, 4): y = 6, z = 7
Case 2: x = 2
The equation becomes 2yz = 3(2 + y + z) or 2yz - 3y - 3z = 6.
This factors as (2y - 3)(2z - 3) = 21 with possible factor pairs (1, 21) and (3, 7).
(1, 21): y = 2, z = 12
(3, 7): y = 3, z = 5
The solutions (x,y,z) that are consistent with the problem constraints are (1,4,15), (1,5,9), (1,6,7), and (2,3,5).
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Posted by xdog
on 2025-02-15 11:40:26 |
solution