Find the number n such that the following alphanumeric equation:

   KYOTO
   KYOTO
 + KYOTO
   TOKYO

has a solution in the base-n number system.

(Each letter in the equation denotes a digit in this system, and different letters denote different digits)

(In reply to FULL solution (part 2) by John Reid)

Foremost, I'd like to say that I think this is one of the most elegant problems that I have come along in quite some time. Not only are the words that make up the equation, Kyoto (kē-ō'tō) and Tokyo (tō'kē-ō'), closely related to each other, but they are made of the same syllables (and spellings) rearranged with different inflections. More info on WORD HISTORY can be found on Answers.com.

We know that:
  n = 9,
  O = 0,
  3 * OT = KY     (3 * [ 9{0} + 1{T} ] = 9{K} + 1{Y} ),
  3 * KY = TO     (3 * [ 9{K} + 1{Y} ] = 9{T} + 1{0} ),
  9 * OT = TO, and
  There is no carry from column 3 to 2.

Since K ≠ 0, T ≠ 1,2,3.

Since Y ≠ 0, T ≠ 0,3,6.

This leaves 4,5,7,8 as possible values of T.

All that is necessary is to check that 3 * T ≠ xL,Lx,LL where L = 0,T.

In base 9:
  3 * 4 = 13  KYOTO is 13040
  3 * 5 = 16  KYOTO is 16050
  3 * 7 = 23  KYOTO is 23070
  3 * 8 = 26  KYOTO is 26080

T, 3T, and 9T are all expressed within KYOTO. What a great problem!

Posted by Charley on 2005-05-14 06:18:48