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)

O must be 0 in this equation because 3O=O can't carry anything. If it did, then O would carry 1 (it can't carry anything more than 2, because if C is what you carry and B is the base (n here) 3O=O+CB, 2O=CB, and since O needs to be less than B, 2 must be greater than B)

Then 3O=O+B, 2O=B, B/2=0. Remember this for later... 3O=K, and since 3O=O without any carried digit added in, K must be greater than O. 3K=T, but if O=B/2 (as stated above), 1.5B+number=T, and T doesn't exist in base B. This is a contradiction, so O must equal 0.

Now you get the following equations equations: 3T=KB+Y, 3Y=CB (since 0+some number can't possibly carry) C+3K=T (C is the same number in these two equations)

Substituting these equations we get 9T=3KB+CB or 9T=(3K+C)B, but since 3K+C=T, 9T=TB, or 9=B if T is nonzero, which it must be since all letters denote different numbers.

Posted by Gamer on 2003-09-21 14:30:18