Please Give Me a detailed solution for this Question .
Find all triples of positive integers (x,y,z) satisfying 99x+100y+101z = 2009 .
99x+100y+101z=2009mod both sides by 100z−x=9z=x+999x+100y+101(x+9)=2009200x+100y+909=2009200x+100y=11002x+y=11y=11−2x
Now y>0 so 1≤x≤5thus we have the following 5 triplets(x,y,z)=(1,9,10), (2,7,11), (3,5,12), (4,3,13), (5,1,14)
Can you please explain how did that modulus function worked .
PLEASE
x mod 100 is just the remainder of x divided by 100
(99x+100y+101z)=100x−x+100y+100z+z100(x+y+z)−x+zwhen we mod by 100 the first part goes away since it's a multiple of 100 which leaves100(x+y+z)−x+z(mod100)=−x+z
How did you got the idea to divide it by 100 .
+296 
