There's A Maximum?

Number Theory Level 5

$\begin{cases} xy+yz+xz+2(x+y+z)=53 \\ x(yz-1)=y+z-2 \end{cases}$

Given that $x,y$ and $z$ are positive integers satisfying the system of equations above, find the maximum value of $x+y$ .

1 solution

Norwyn Kah
May 29, 2016

We can get that xyz=x+y+z-2

Using the identity (x+1)(y+1)(z+1)=xyz +xy+xz+yz+x+y+z+1

Substituting you'll get that (x+1)(y+1)(z+1)=(2x+2y+2z+xy+xz+yz)-1=53-1

(x+1)(y+1)(z+1)=52 WLOG we can get the solutions (51,0,0) (25,1,0) (12,1,1) (12,3,0)

Since x,y,z are positive integers the only solution is (12,1,1) finding the maximum value for x+y will be 12+1=13

Same approach 👌

Grant Bulaong - 5 years ago