Problem 4

Number Theory Level 2

How many ordered triplets ( x ; y ; z ) (x;y;z) of non-zero positive integers exists to: 2 x + 3 y + 4 z = 17 2x+3y+4z=17 ?


The answer is 4.

Victor Paes Plinio by Victor Paes Plinio , 21, Brazil

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3 solutions

Josh Speckman
Jul 24, 2014

1 1 3

2 3 1

3 1 2

5 1 1

Also, nonzero positive integers is redundant

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Since 17 - 3y has to be divisible by 2, y can only be 14, and 8.(y=2 makes z=0!)
Since there has to be at least one x, z can only take from 14-2=12 and 8-2=6.
So there can be only 12/4 = 3 & 6/4 = 1...".z"..> 3 + 1 = 4 3 + 1 = 4

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Raghav Mehta
Jul 9, 2014

put values for z to be 1,2,3,4 and then solve for x and y

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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