Let $N$ be a six digit number with distinct digits. We have to find the number $N$ given that the numbers $N,\hspace{0.5em}2N,\hspace{0.5em}3N,\hspace{0.5em}4N,\hspace{0.5em}5N$ and $6N$, when written underneath each other, form a latin square (that is each row and each column contains all six digits).

Let $N=\mathrm{abcdef}={10}^{5}a+{10}^{4}b+{10}^{3}c+{10}^{2}d+10e+f$ and $S=a+b+c+d+e+f$. Then

${\displaystyle 21N=N+2N+3N+4N+5N+6N=S\times 111111}$

Therefore

${\displaystyle N=S\times 5291.}$

Now $S\ge 1+2+3+4+5+6=21$. As $6N$ has only 6 digits it follows that $a=1$. Hence $N\le 198765$ and $S\le 198765/5291\le 37$. We now check by computing $N=5291S$ for $21\le S\le 37$ 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: