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: