Solving for integer solutions!

Number Theory Level 3

Let x , y x,y and z z be positive integers satisfying

2 x y z = x + y + z . 2xyz = x+y+z.

How many ordered triplets ( x , y , z ) (x,y,z) satisfy the equation above?


The answer is 3.

Tin Le by Tin Le , 15, Vietnam

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Tin Le
Aug 9, 2020

Without the loss of generality, assume that 1 ≤ x ≤ y ≤ z 1 \leq x \leq y \leq z .

Therefore, 2 x y z = x + y + z ≤ 3 z ⇔ 2 x y ≤ 3 2xyz = x+y+z \leq 3z \Leftrightarrow 2xy \leq 3

⇒ x y ≤ 1 ⇒ x y = 1 ⇒ x = y = 1 \Rightarrow xy \leq 1 \Rightarrow xy = 1 \Rightarrow x=y=1 .

Substituting x = y = 1 x=y=1 into the equation, we have 2 z = 2 + z ⇔ z = 2 2z=2+z \Leftrightarrow z=2

As all cases are all the same, there are 3 \boxed{3} solutions, which are ( x , y , z ) = ( 1 , 1 , 2 ) , ( 1 , 2 , 1 ) , ( 2 , 1 , 1 ) (x,y,z) = (1,1,2), (1,2,1), (2,1,1)

2 Helpful 0 Interesting 1 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...