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 = Kn + 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 theres at least one solution for this n. In fact, there are four: KYOTO = 13040, 16050, 23070, or 26080.