The answer is *n* = 9. Inspecting the first digit from the right in the given
equation, we find that 2 × O is divisible by *n*. So either O = 0 or O = *n*/2.
In the second case, from the third digit we derive K > *n*/2, but from the fifth
digit we see that 3 × K <= [“less than or equal to”] T <= *n* –
1, so K < *n*/3. It follows that O = 0. Now we have these equations: 3 × T = K*n*
+ Y (from the second and third digits), 3 × Y = *cn* (where *c* is the number
carried from the fourth to the fifth digit), and 3 × K + *c* = T (the fifth digit).
Multiplying the first equation by 3 and substituting the expressions for 3Y and 3K from
the two other equations, we get 9T = (T – *c*)*n* + *cn*, and so *n*
= 9. We must also check that there’s at least one solution for this *n*. In
fact, there are four: KYOTO = 13040, 16050, 23070, or 26080.