Answer to equation
Key to colouring:  Newly worked out – that line, Previously worked out. 
DONALD +GERALD =ROBERT 

5ONAL5 +GERAL5 =ROBER0 1 
Given that D = 5 Therefore T = 0 (5 + 5 = 10) 
5ONAL5 +G9RAL5 =ROB9R0 11 11 
R must be an odd number because of the 1 carried over (L + L + 1 = R), So R = 1, 3, 7 or 9 Also E = 9 because it is the only number that will allow O to be part of the answer. I.e. any number plus 10 is for the only way it can appear in the answer but to allow this N+R must now equal B+10 so a one gets carried over. 
5ON4L5 +G9R4L5 =ROB9R0 11 11 
If E = 9, A+A cannot equal an odd number (answer must be divisable by [2 A + A = 2xA]) so L+L+1 must equal R +10 (ie a ten has to carried from the tens into the hundreds to make the answer an odd number) 2A+1 can only be = 9 or 19 2A = 9 – 1 or 19 – 1 2A = 8 or 18 but E = 9 so A = 4 only 
5ON4L5 +1974L5 =7OB970 11 11 
5+G+1= R must be less than 10 (nothing carried), therefore G = 0 – 4 (but T = 0 and A = 4) so 1, 2, or 3 when substituted in R = 7, 8 or 9, but R must be an odd number and E = 9 so R = 7 and therefore G = 1. 
5ON485 +197485 =7OB970 11 11 
R=7 so 2L+1=7+10 so 2L=16 –> L=8 
5O6485 +197485 =7O3970 11 11 
N+7 must be greater than 10 ie N = 3 or more [N >=3] also N+7 = B+10 ie N = B+3, What’s left? N cannot be = 4, 5, 7, 8, 9 If B=3 then N=6 if B=6 then N=9, but E=9 so B=3 only 
526485 +197485 =723970 11 11 
1=G, 2=?, 3=N, 4=A, 5=D, 6=N, 7=R, 8=L, 9=E, 0=T Therefore by exclusion O=2. 
526485 +197485 =723970 11 11 
Alternative
sent in by Alan Farrell, University of Texas at Brownsville
[Prerequisite: High School Algebra]
The solution is a bit lengthy and may be a little
difficult to follow if you don’t remember your algebra, but here goes.
First of all, we must assume that each letter represents a unique positive
integer, in other words, two different letters may not represent the same
integer. There are 10 letters and 10 integers (09). Let each
column of letters be denoted the numbers 16 from left to right. Since we
cannot yet determine if a 1 must be carried over to the next column we add a
row above DONALD labeled x1 x2 x3 x4 x5 x6. So we have
X1  X2  X3  X4  X5  X6  
D  O  N  A  L  D  
+  G  E  R  A  L  D 
=  R  O  B  E  R  T 
where each x value is either 0 or 1.
Obviously x6=0 since column 6 is the first column to be added (if this is
confusing you then try adding two 6 digit numbers and you’ll see what I mean).
We are given that D=5, therefore we can
immediately determine that T=0 and that x5=1. So far we have
X1  X2  X3  X4  1  0  
5  O  N  A  L  5  
+  G  E  R  A  L  5 
=  R  O  B  E  R  0 
Next we look at column 2 and note that the sum
x2+O+E=O. Since we do not know if x2 equals 0 or 1, we must consider
both possibilities. If x2=0 and O+E<10 then O+E=O and E=0, but T=0 so
this cannot be true. If O+E>10 then O+E=O+10 and E=10, but
our variables must take values from 09 so this cannot be true. Now
we know that x2=1 since we eliminated the possibility x2=0. With x2=1,
if 1+O+E<10 then 1+O+E=O, or O+E=O1 and E=1, but this is negative so
this cannot be true. If 1+O+E>10 then 1+O+E=O+10, or O+E=O+9 and E=9.
So now we have:
X1  1  X3  X4  1  0  
5  O  N  A  L  5  
+  G  9  R  A  L  5 
=  R  O  B  9  R  0 
From column 4 we have x4+A+A=9. We cannot have x4=0 since then 2A=9
and A=4.5, which is not an integer. Thus, x4=1 and 1+2A=9, which
gives A=4, and x3=0. So now we have
X1  1  0  1  1  0  
5  O  N  4  L  5  
+  G  9  R  4  L  5 
=  R  O  B  9  R  0 
Next we look at column 1 and note that the sum
x1+5+G=R. It follows that x1+G=R5 and thus 5<=R<=9 (since x1+G
must be positive). Now that we have a little information about R, we
look at column 5 and note that the sum 1+L+L=1+2L=R+10 (because x4=1) .
This tells us that R is an odd number (2 times any number plus 1 is odd, try
it if you don’t believe me). So R may only take the value 5, 7 or 9, but
5 and 9 are already taken, so R=7. Since 1+2L>10 we have 1+2L=10+R,
or 1+2L=17 and we find that L=8.So now we have
X1  1  0  1  1  0  
5  O  N  4  8  5  
+  G  9  7  4  8  5 
=  7  O  B  9  7  0 
Since column 2 is greater than 10, x1=1. Then from column 1 we get
1+5+G=7, thus G=1. We are now left with the variables O, N, and B,
and the integers 2, 3, and 6. Column 3 gives N+7=B+10, or NB=3.
Thus, N=6 and B=3. Since 2 is the only integer left, we must have
O=2. Finally we have
1  1  0  1  1  0  
5  2  6  4  8  5  
+  1  9  7  4  8  5 
=  7  2  3  9  7  0 
We can now write this without the carried 1’s and get
5  2  6  4  8  5  
+  1  9  7  4  8  5 
=  7  2  3  9  7  0 