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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
avatar+5095 
+1

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

 

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

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 Nov 8, 2018
 #2
avatar+294 
+1

Can you please explain how did that modulus function worked .

PLEASE

Darkside  Nov 8, 2018
 #3
avatar+5095 
+1

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
 #4
avatar+294 
+1

How did you got the idea to divide it by 100 .

Darkside  Nov 8, 2018

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