help

Let's define a few variables to start.

 

Let A = one digit of the three-digit number

Let B = one digit of the three-digit number

Let C = one digit of the three-digit number

 

For problems like these, it is generally helpful to find an expression that relates everything together.  \(ABC, ACB, BAC, BCA, CAB, CBA\) represent all the six different 3-digit numbers. We can then represent the sum of all the unique 3-digit numbers with the expression \(222(A+B+C)\) because each digit is added twice in the units, tens, and hundreds digit. 

 

Also, it is generally helpful to find a bound of some kind that restricts the possibilities. This way, it is not necessary to check every single possibility mindlessly. We can achieve this by recognizing that we can represent the sum in a different way. The sum must be at least \(3231+123=3354\) and must not exceed \(3231+987=4218\).

 

With this information, we can solve an inequality that restricts the possibilities for \(A+B+C\).

\(3354\leq 222(A+B+C)\leq4218\\ 15.108...\leq A+B+C \leq 19\)

 

Of course, in the context of this problem, \(A+B+C\) is restricted to the set of whole numbers. Therefore, \(A+B+C\) equals 16,17,18, or 19. After this point, we will just test all the possibilities since we only have 4 to check anyway.

 

We can represent the 6th number as \(222(A+B+C)-3231\). Below, I have included a table for each possibility for \(A+B+C\):

 

\(A+B+C\)	6th Number
\(16\)	\(321\)
\(17\)	\(543\)
\(18\)	\(765\)
\(19\)	\(987\)

 

From this table, only one of possibilities for the 6th number reigns supreme, 765, because 765 is the only number where its digits add to its corresponding \(A+B+C\) value.