This is Q27 from Australian Maths 2013.
$3$ different non-zero digits are used to form $6$ different $3$-digit numbers.The sum of $5$ of them is $3231$.What is the $6$ th number?
What I tried:
Let $a,b,c$ be the different digits.
$(100a+10b+c)+(100a+10c+b)+(100b+10a+c)+(100b+10c+a)+(100c+10a+b) =3231 $
From there,I can see that
$a+2b+2c =10x +1 $,where $x$ is some integer.
$2a+b+2c =10j+(3-x)$,where $j$ is some integer.
Using substitution to sub in the values of $j$ and $x$,
$221a +212b+122c=3231$,which leads me back to where I started from...