## Friday, June 29, 2007

### Donald, Gerald, and Robert

I learned a lot of things during my recent trip to Minneapolis, but the most important was how to solve logic puzzles. The fact that I can now tackle a reasonably difficult puzzle is pretty striking, since I used to be hopeless at them.

So when I ran across this classic I just HAD to try and solve it, and I'm proud to say it only took me...errr, fifteen minutes. That might not be so hot but I'M pretty pleased with myself.

"DONALD + GERALD = ROBERT" is a mathematical cryptogram. Each of the ten letters stands for one of the digits from 0 to 9. None of the digits stand for more than one letter. As a starter, the letter "D" stands for "5".

For those of you (like me) who hate puzzles that require trial and error to solve, don't worry; this one is unambiguous.

Eric Little said...

I look at those names and think, "Duck + Ford = Frost."

thinkulous said...

Pretty good, Eric!

Yeah, have to admit, math puzzles make my spine shrink.

Hats off to you, though, Muffy, for mastering them as much as you have!

Alan said...

Solve: D O N A L D + G E R A L D = R O B E R T, D = 5, each letter represents a unique digit from 0 to 9.

_
_ D O N A L D
+ G E R A L D
_ == == == == == ==
_ R O B E R T

1. Since D = 5, and D + D = T, then 5 + 5 = 10, so T is 0, and we carry a 1:
_ _ _ _ _ _ 1
_ 5 O N A L 5
+ G E R A L 5
_ == == == == == ==
_ R O B E R 0

2. Since O + E = O, and E cannot be zero (0 is already taken), this addition can only be possible if E = 9. Then, O + 9 = O is only possible if we have a carry-over of 1 from the addition to its right, (N + R = B).
We observe that O + 9 = O, which means that the result is > 10, and we must have a carry-over of 1. We have:
_ _ 1 _ _ _ 1
_ 5 O N A L 5
+ G 9 R A L 5
_ == == == == == ==
_ R O B 9 R 0

3. Since A + A = 9, we must have a carry-over from the previous column (from column L + L = R), and A must equal to 4 (i.e. 4 + 4 + 1 = 9). Thus, A=4.
_ _ 1 _ _ 1 1
_ 5 O N 4 L 5
+ G 9 R 4 L 5
_ == == == == == ==
_ R O B 9 R 0

4. Since we carried over a 1 to accomplish O + 9 = O, then N + R = B implies that B > 10 (we’re carrying the 1 over, and 0 is already taken by T, and that is why B cannot equal 10, so B > 10)

Digits used up: { 0, 4, 5, 9 } Free: { 1, 2, 3, 6, 7, 8 }
Since N + R > 10, then from the Free set we can only try to use combinations {3, 8}, {6, 7}, {6, 8}, or {7, 8}.
We also observe that L + L + 1 = R, which implies that R must be odd. Then, from our 4 sets of possible assignments above, we can only use 3 and 7 as possible assignments to R, since they are the only odd digits occurring in the sets. Thus, R must be either 3 or 7.

We now try to figure out if R is 3 or 7.
In the first column we see that 5 + G + 1 = R (and since we do not have any more digits left of R, i.e. we don’t carry over anything), R cannot be 3 (since 5 + G + 1 = 3 implies that G is negative).

Thus, R must take value 7. We have:

_ _ 1 1 _ 1 1
_ 5 O N 4 L 5
+ G 9 7 4 L 5
_ == == == == == ==
_ 7 O B 9 7 0

5. Since 5 + G + 1 = 7, then G = 1. We have:
_ _ 1 1 _ 1 1
_ 5 O N 4 L 5
+ 1 9 7 4 L 5
_ == == == == == ==
_ 7 O B 9 7 0

6. Since L + L + 1 = 7, then we must have L = 3 or L = 8.
We notice from 4 + 4 = 9 that we must be carrying a 1 over from L + L.
Since we carry a 1 from L + L, then we cannot have L = 3, and therefore L = 8:
_ _ 1 1 _ 1 1
_ 5 O N 4 8 5
+ 1 9 7 4 8 5
_ == == == == == ==
_ 7 O B 9 7 0

7. Digits used up: {1, 4, 5, 7, 8, 9, 0} Free: {2, 3, 6}
We try to see which free digits match next.
We have N + 7 = B, and we know that we must carry a 1 after the addition, thus we can’t have N=2, because 2 + 7 < 10. If we let N = 3, then 3 + 7 = 10, and B cannot be zero (zero is already taken), therefore N cannot equal 3. The only other possibility is for N = 6.
It immediately follows that B = 3 (since 6 + 7 = 13)
_ _ 1 1 _ 1 1
_ 5 O 6 4 8 5
+ 1 9 7 4 8 5
_ == == == == == ==
_ 7 O 3 9 7 0

8. The only remaining unused digit is 2. Therefore, let O = 2. Then 2 + 9 + 1 = 12, which works out. The final equation is then:
_ _ 1 1 _ 1 1
_ 5 2 6 4 8 5
+ 1 9 7 4 8 5
_ == == == == == ==
_ 7 2 3 9 7 0

Adding 526485 with197485 gives the result: 723970, therefore the addition is correct. The mapping is thus:
0 = T, 1 = G, 2 = O, 3 = B, 4 = A, 5 = D, 6 = N, 7 = R, 8 = L, 9 = E.

(2008, by Alan Lupsha)

Polo said...

Good Eric!
I have just read the puzzle (to take hint :-))and it's interesting because a grandfather has sent it to me!!