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Old 10-16-2002, 12:20 PM
Homer Homer is offline
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Join Date: Sep 2002
Posts: 5,909
Default Re: Bartlett\'s problem

Ok, here we go...this was a good one.

My answers are:

T = 0
G = 1
O = 2
B = 3
A = 4
D = 5
N = 6
R = 7
L = 8
E = 9

So that makes the solution:

526485 (DONALD) +
197485 (GERALD) =
723970 (ROBERT)

Here is how I went about solving it:
1) We are given D = 5
2) D + D = 10, so T = 0 (and we carry the one)
3) 2L + 1 = R
- We know that R >= 6 since 5 + G = R (and G cannot equal 0)
- We also know that R is odd based on the 2L + 1 = R equation.
- Therefore, R is either 7 or 9
4) E is equal to either 0 or 9, because O + E = O or O + E + 1 = 10 + O
- Since 0 is already taken, E = 9
- Since E is 9, and R is either 7 or 9, R = 7 by default since 9 is taken by E
5) A + A = 9 or 19. For this to be true, we know that a one must be carried to that column from the previous column, making A either 4 or 9. It can't be 9 because it is already taken, so A = 4.
6) In the previous column from A + A = E, we have L + L + 1= 7 + 10 (add the ten because we know from step 5 that a 1 will have to be carried). Thus, L = 8
7) N + R = B. N + R > 10, because we are going to have to carry a 1 from that column to the next, since E = 9, to make that equation O + 9 + 1 = O + 10. We know R = 7, so N > 3. The only remaining number greater than 3 is 6. So, N = 6
8) Since N = 6, N + R = 13, making B = 3.
9) D + G + 1 = R, and D = 5 and R = 7. So, G = 1.
10) We are left with O=2

....I didn't explain myself well, I am sure, but I don't know what I am thinking half the time, so it is hard to get it down on paper. By the way, this took me about 15 minutes. I am guessing that others may be able to do it faster.

-- Homer
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