$5$ numbers add up to 3231.What is the $6$th number? 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.
$2a+2b+c =32-j$
Using substitution to sub in the values of $j$ and $x$,
$221a +212b+122c=3231$,which leads me back to where I started from...

With three digits $a,b,c$, You should be able to get at most six different $3$-digits numbers, and they are: $abc,acb,bac,bca,cab,cba$So when you add them up, the equation should be$$200(a+b+c)+20(a+b+c)+2(a+b+c)=222(a+b+c)=3231+n$$where $n$ is the unknown $6$th number. 
By quick estimation you can find that, 
When $a+b+c=15, n=99$, does not qualify;
When $a+b+c=16, n=321$, does not qualify;
When $a+b+c=17, n=543$, does not qualify;
When $a+b+c=18, n=765$, qualified.
