Substitute digits for the letters to make the following relation true.

        S O M E 
      + G O O D 
     ---------- 
      I D E A S 
     ---------- 

Note that the leftmost letter can't be zero in any word. Also, different digits represent different values.

(In reply to re: A solution by fwaff)

I know this is an old problem, and i know that Fwaff explained some of his approach to the correct answer, but I didn't feel like any comments, or the official solution, fully explained how to arrive at the solution.  So here goes:

I must be 1 because even 9999+9999 = 19998, so we’ll never get to 2xxxx.

E+D may or may not have a carry over, so we either have E+D = S or E+D = 10+S.
We know S+G will have a carry over (because of the I) so we know the sum will equal 10+D. However, we don’t know if there will be a carry over from O+O. So we either have S+G = 10+D or S+G+1 = 10+D

S+G = E+D+G = 10+D.
So E+G = 10 but that would make S be 0 which isn’t possible since it is also a leading digit.

S+G = E+D-10+G = 10+D.
So E+G = 20 which isn’t a possible sum for two single digit numbers.

S+G+1 = E+D+G+1 = 10+D.
So E+G = 9

S+G+1 = E+D-10+G+1 = 10+D.
So E+G = 19 which isn’t possible for single digit numbers.

Therefore E+G = 9, E+D=S has no carry over (E+D<9), and S+G+1 = D+10 so O+O must have a carry over (O>=5).

S can’t be:
0 since it leads the word SOME
1 since that is taken by I
2 because then E=D=1=I (no 5+7 wouldn’t work because E+D has no carry over)
3 because then E=1=I and D=2 (or vice versa)
4 because then we either have E+D equals 1+3 or 2+2
9 because then D=G (E+G = 9 = S = E+D)

S can only be 5 (2+3), 6 (2+4), 7 (2+5 or 3+4) or 8 (2+6, 3+5). This will force the value for G as well since G=9-E. D never has a contradiction. There is one case where G=S, so I’ll eliminate that (S=7, E=2, D=5, G=9).

However, based on these possibilities for E, this forces O to be a certain number. E = 2,3,4,5,6.
If E is even, that means that M+O did not have a carry over. Remember, O+O does have a carry over, so in this case O+O=10+E, or O = (10+E)/2.
If E is odd, that means M+O did have a carry over. So in this case O+O+1 = 10+E, or O = (9+E)/2.

So, when E=2, O=6. When E=3, O=6. When E=4, O=7. When E=5, O=7. When E=6, O=8. But in some of those cases, either S, E or G equals O which is not allowed. So we only have the following allowable cases:

S=5, E=2, D=3, G=7, O=6.
E is even so M+O can’t have a carry over, so M<4 so M=0 since 1,2,3 are taken. But this forces A=6=O.  So no.

S=6, E=4, D=2, G=5, O=7.
E is even so M+O can’t have a carry over, so M<3 so M=0 since 1,2 are taken. But this forces A=7=O. So no.

S=8, E=5, D=3, G=4, O=7.
E is odd so M+O must have a carry over, so M>2 so M = 6 or 9. M=6 makes A=3=D. M=9 makes A=6 which is free.

So the answer is

  8795
+ 4773
------
 13568

Posted by nikki on 2005-01-27 16:37:27