QUOTE(Victoria Silverwolf @ Oct 15 2003, 02:46 PM)
Here's a cryptarithm I made up a long time ago, which was published in The Journal of Recreational Mathematics. Each letter in these two equations represents a digit from 0 to 9. The letters are consistent in both equations; that is, the "O" in "TWO" represents the same digit as the "O" in "FOUR" and so on.
ONE + ONE = TWO
ONE + FOUR = FIVE
There are two possible solutions.
I think I got it!
0 = R
1 = V
2 = O
3 = N
4 = T
5 = I
6 = E
7 = W
8 = U
9 = F
so
ONE + ONE = TWO
is
236 + 236 = 472
and
ONE + FOUR = FIVE
is
236 + 9280 = 9516
Whoo hoo!

edited to add explanation/process:
First, R must equal zero because e+e=o AND e+r=e
Then, O must be an even number because e+e=o
Next, T must be 1 plus or minus I because o+o=t and o+o=i
Next, N or U must be greater than 5 because n+u or n+n must be greater than 10 (for o+o=t and o+o=i)
Finally, O must either be 2 or 4 because T cannot be greater than 9 (e+e=o and o+o=t)
I think those are the main steps I took to get to the solution.