Australian mathematics competition

Let the three different non-zero digits be $a$ , $b$ , and $c$ . The three digits are even distributed in the six numbers that they form. That is each of them appears twice as the hundreds digit, twice as the tens digit, and twice as units digit in the six numbers. Therefore, the sum of the six numbers is $222(a+b+c)$ or $222s$ , where $s=a+b+c$ , the sum of digits.

$\begin{array} {ccc} a & b & c \\ a & c & b\\ b & a & c\\ b & c & a\\ c & a & b\\ c & b & a \end{array}$

Now, let us estimate the sum of digit $s$ . $\frac {222s}6 \approx \frac {3231}5 \implies s \approx \frac {3231}5 \times \frac 6{222} \approx 17.46$ . Therefore $s$ is likely to be $17$ or $18$ . Since $3231$ is divisible by $9$ as its digital sum is divisible by $9$ , $s$ is more likely to be $18$ , which is also divisible by $9$ .

To find the sixth number, we just use $n_6 = 222s-3231$ and for $s=18$ we have $n_6= 3996 - 3231 = 765$ . And we note that $n_6 = 765$ has a digital sum of $18$ . Therefore our assumption is correct and $\boxed{765}$ is the sixth number. Using other $s$ do not give the correct result.

Let $a,$ $b$ , and $c$ be the digits. The sum of the 6 permutations of these digits is $222(a + b + c)$ . Without loss of generality, $a < b < c$ .

As a first step, I consider the various multiples of $222$ :

$\{222*n : 1 \leq n \leq 9\} = \{222, 444, 666, 888, 1110, 1332, 1554, 1776, 1998\}.$

Clearly $c$ cannot be a low digit, since $222(a + b + c) \leq 222(3c - 3)$ , but $222(a + b + c) > 3231$ .

We can quickly dismiss $c \leq 6$ , as the maximum sum subject to this constraint is $222(4 + 5 +6 ) = 888 + 1110 + 1332 = 3330$ , but $3330 - 3231 = 99$ , which is too small.

My general approach will be to search over triples $a,b,c$ by considering the expression $222(a + b + c) - 3231$ and testing to see if this difference is a three-digit permutation of the digits $a, b$ , and $c$ .

With $c \geq 7$ , in theory I only have $\binom{6}{2} + \binom{7}{2} + \binom{8}{2} = 15 + 21 + 28 = 64$ triples to search!

My search starts with $c = 7$ . To maximize $a + b + c$ , I let $a = 5$ and $b = 6$ and I get lucky!

$222*(5 +6 + 7) = 3996$ and $3996 - 3231 = 765$ , which is a permutation of the three chosen digits.

Found the answer relatively quickly. Not promising no other solutions exist.

Each digit appears twice in the hundreds, the tens and the units column amongst the six numbers which can be formed. If the digits are a, b, c and the missing number is c, then 3231 + — 222(a + b + c). Then = 222m — 3231 must have three distinct digits that add to m. So m > 3231/222 = 15B. Try m — 16, 17, . . .. 222 x 16 - 3231 — 3552 - 3231 = 321, 3 +2 + 1 16 222 x 17 - 3231 — 3774 3231 — 543, = 12 17 222 x 18 - 3231 — 3996 - 3231 = 765, So the sixth number is 765, 7+6+5 = 18
hence (765).