Guest Posted November 28, 2009 Report Share Posted November 28, 2009 Determine all possible triplet(s) (x, y, z) of positive integers, with x ≤ y ≤ z, that satisfy this equation: 1/x + 1/y + 1/z = 3/13 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 28, 2009 Report Share Posted November 28, 2009 Well, it can only be (13,13,13) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 28, 2009 Report Share Posted November 28, 2009 x=10. y=10, z=65/2 there are more than just one solution Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 28, 2009 Report Share Posted November 28, 2009 They have to be integers, M. Try the following triplets :- (x,y,z) = (5, 33, 2145), (5, 35, 455), (5, 39, 195), (5, 45, 117), (6, 16, 624), (6, 18, 117), (6, 26, 39), (7, 13, 91). Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 28, 2009 Report Share Posted November 28, 2009 (edited) 9, 9, 117 also the relation between x, y, and z in 1/y+1/x+1/z=3/13is z=13xy/(3xy-13x-13y) 1/x+1/y+1/z=3/13 subtract 1/x+1/y from each side 1/z=3/13-1/x-1/y raise both sides to the -1 power (1/z)-1=(3/13-1/x-1/y)-1 simplify exponents z=1/(3/13-1/x-1/y) multiply the right side of the equation by (13xy)/(13xy) z=13xy/[13xy(3/13)-13xy(1/x)-13xy(1/x)] then simlify z=13xy/(3xy-13x-13y) i haven't been able to figure out how to work in the x<y<z Edited November 28, 2009 by mortsemious Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 28, 2009 Report Share Posted November 28, 2009 Determine all possible triplet(s) (x, y, z) of positive integers, with x ≤ y ≤ z, that satisfy this equation: 1/x + 1/y + 1/z = 3/13 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 28, 2009 Report Share Posted November 28, 2009 1,1,11 1,2,10 1,3,9 1,4,8 1,5,7 1,6,6 2,2,9 2,3,8 2,4,7 2,5,6 3,3,7 3,4,6 3,5,5 4,4,5 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 29, 2009 Report Share Posted November 29, 2009 5 ---- 33 ---- 2145 5 ---- 35 ----- 455 5 ---- 39 ----- 195 5 ---- 45 ----- 117 6 ---- 16 ----- 624 6 ---- 18 ----- 117 6 ---- 26 ------ 39 7 ---- 13 ------ 91 9 ----- 9 ------ 117 Quote Link to comment Share on other sites More sharing options...
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Determine all possible triplet(s) (x, y, z) of positive integers, with x ≤ y ≤ z, that
satisfy this equation:
1/x + 1/y + 1/z = 3/13
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