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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 .

Nov 8, 2018

#1
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$$99x+100y+101z = 2009 \\ \text{mod both sides by 100}\\ z-x = 9 \\ z=x+9\\ 99x + 100y + 101(x+9) = 2009 \\ 200x + 100y + 909 = 2009\\ 200x + 100y = 1100\\ 2x+y=11\\ y=11-2x$$

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

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Nov 8, 2018
#2
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Can you please explain how did that modulus function worked .

Darkside  Nov 8, 2018
#3
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x mod 100 is just the remainder of x divided by 100

$$(99x + 100y + 101z) = \\ 100x - x+100y + 100z + z \\ 100(x+y+z) - x + z\\ \text{when we mod by 100 the first part goes away since it's a multiple of 100 which leaves}\\ 100(x+y+z) - x + z \pmod{100} = -x + z$$

Rom  Nov 8, 2018
edited by Rom  Nov 8, 2018
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How did you got the idea to divide it by 100 .

Darkside  Nov 8, 2018