All possible values of N (3-digit numbers) possible formed by non-zero digits

Question

3 different non-zero digits are chosen to make 6 different 3-digit numbers, with no digits repeated in any number. Some of the 3-digit numbers, N say, is the average of the other five. Find all possible values of N.

Answer 1

Rather than go by combination, one may choose to use variables to express the given condition.

Given $a,b,c$ non-zero pairwise distinct, the six numbers that can be formed are $100a+10b+c,100b+10a+c, $etc. (just permute $a,b,c$ in any of six possible orders).

Now, if, for example, $100a+10b+c$ is the average of the other numbers, then five times $100a+10b+c$ is the sum of the other numbers. We know what the other numbers are, so let's write this down: $$ 5(100a+10b+c) = (100b+10a+c)+(100c+10b+a)+(100b+10c+a) + (100c+10a+b)+(100a+10c+b) $$

this simplifies to: $$500a+50b+5c = 212b+122a+221c \implies 378a=162b+216c $$ miraculously, $54$ can be cancelled from both sides of the above equation, and this leaves us with $7a=3b+4c$.

So the question is this:which distinct single digit $a,b,c$ satisfy $7a=3b+4c$?

What helps in our simplification of cases is that transposing in the above equation gives $3(a-b) = 4(c-a)$ (you can expand and check that this is the same as $7a=3b+4c$).

What this means, is that $a-b$ is a multiple of $4$ and $c-a$ is a multiple of $3$. Furthermore, either $c>a>b$ or $b>a>c$ must happen. That is, $a$ is always in the middle.

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Now, fix some value for $a$. Find all $b,c$ suitable and check if the equation above is satisfied.

$a=1$ is not possible : it has to be in the middle.

$a=2$ forces either $b$ or $c = 1$, which contradicts the condition we have. Similarly, $a=3$ does not work.

$a=4$ gives $c = 1$ and $b = 8$. So, $481$ is the average of the five other permutes of this number!

$a = 5$ gives $b=9,c=2$ or $b=1,c=8$, so $592,518$ are two other numbers.

$a=6$ gives $b=2,c=9$ so $629$ is another number.

You can check that no other possibilities are there. Interestingly enough, $$ 481 = 37 \times 13 \\ 518 = 37 \times 14 \\ 592 = 37 \times 16 \\ 629 = 37 \times 17 $$

you can try to find why this happens!