Well there are 10 letters, ADEGJOPRSY, and I assume it's in decimal. Typically, each letter is unique. If it's in hexadecimal or a higher base, then it becomes much more difficult. You can think A has 10 possible values, D has 9 possible values, E has 8, and as a result there are 10! or 3628800 possibilities, assuming it's base 10 arithmetic.
There is some information though, for example G+P+carry = E <= 9, carry either being 0 or 1. As a result, G and P are not 9, possibly not 8. Each number is four letters/digits long, and I would focus on the carries. Each digit either has a carry or not a carry, and that gives 8 possibilities.
Take the least significant digits, D+J = Y. If there is no carry, then 0 < D,J < Y <= 9. D and J must be nonzero, otherwise D = Y or J = Y. That hinges on the uniqueness requirement. Also, 2*O = S. Otherwise, 0 <= Y < D,J <= 9 and O+O+1 = S.
If the problem is in decimal, then there must be at least one carry that causes a result to be zero. Also, it's probably easier to not consider each carry, because it may lead to duplicates. While you can prune some options, after that I would greedily select whichever number has the most possibilities (probably 5).
