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Thank you for this solution to Andrei Lazanu, School 205,
Bucharest, Romania and Jonathan Smith, Gresham's School,
Norfolk.
Let $N$ be a six digit number with distinct digits. We have to
find the number $N$ given that the numbers $N,\ 2N,\ 3N,\ 4N,\ 5N$
and $6N$, when written underneath each other, form a latin square
(that is each row and each column contains all six digits).
Let $N = abcdef = 10^5a + 10^4b +10^3c + 10^2d + 10e + f$ and
$S = a + b + c + d + e + f$. Then $$21N = N + 2N + 3N + 4N + 5N +
6N = S \times 111111$$
Therefore $$N = S \times 5291.$$
Now $S \geq 1+2 + 3 + 4 + 5 + 6 = 21$. As $6N$ has only $6$
digits it follows that $a=1$. Hence $N \leq 198765$ and $S \leq
198765/5291 \leq 37$. We now check by computing $N = 5291S$ for $21
\leq S \leq 37$ and also $2N$, and visually check the digits of $N$
and $2N$ to see if they are the same. The solution, written as a
latin square, is:
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Published September 2002.