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Please Give Me a detailed solution for this Question .

Find all triples of positive integers (x,y,z) satisfying 99x+100y+101z = 2009 .

Darkside Nov 8, 2018

#1**+1 **

\(99x+100y+101z = 2009 \\ \text{mod both sides by 100}\\ z-x = 9 \\ z=x+9\\ 99x + 100y + 101(x+9) = 2009 \\ 200x + 100y + 909 = 2009\\ 200x + 100y = 1100\\ 2x+y=11\\ y=11-2x\)

\(\text{Now }y>0 \text{ so }1\leq x \leq 5\\ \text{thus we have the following 5 triplets}\\ (x,y,z) = (1,9,10),~(2,7,11),~(3,5,12),~(4,3,13),~(5,1,14)\)

.Rom Nov 8, 2018