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 (0-9).  Let each
column of letters be denoted the numbers 1-6 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 0-9 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=O-1 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=R-5 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 N-B=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
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