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 = 10⁵a + 10⁴b +10³c + 10²d + 10e + f and S = a + b + c + d + e + f. Then

21N = N + 2N + 3N + 4N + 5N + 6N = S ×111111

Therefore

N = S ×5291.

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