|Key to colouring:||Newly worked out – that line,
Previously worked out.
|Given that D = 5
Therefore T = 0 (5 + 5 = 10)
|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
|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
|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.
|R=7 so 2L+1=7+10
so 2L=16 –> L=8
|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
|1=G, 2=?, 3=N, 4=A, 5=D, 6=N,
7=R, 8=L, 9=E, 0=T
Therefore by exclusion O=2.
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
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
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:
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
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
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
We can now write this without the carried 1’s and get