Bartlett's problem

[happyjaypee]

This one was created in 1958 so for those who know it, don't spoil the problem by giving the answer right away pls.

Can you solve this math problem?:

DONALD+GERALD=ROBERT

Assuming D=5

Add the process you went trought whit the answer.

2 more things:
All letters have a value of 0 to 9
Two differents lettres can't have the same value

[happyjaypee]

[lorinda]

I dont know this one and will work it out including workings, which may be interesting for those who know the answer.
I shall also wait before posting for a couple of days.
I shall also probably mess it up.

[PseudoPserious]

t=0, g=1, o=2, b=3, a=4, d=5, n=6, r=7, l=8, e=9

526485 + 197485 = 723970

d=5: given

t=0: from the 1's place, 5+5=10

e=9: from the 10,000's place, we have o+e=o. Since e is not 0, this must be 1+o+e=o+10 (that is, we have a 1 carrying over from the 1,000's place). Cancelling, e=9.

a=4: from the 100's place, a+a=e. Since e=9, we have to have a carry over from the 10's place. So, either 1+a+a=19 or 1+a+a=9. As a can't be 9, we have 1+4+4=9 and a=4.

r=7: from the 100,000's place, 1+5+g=r (we have a carry over from o+e=o). So, r is 7 or 8 and g is 1 or 2. From the 10's place, we have 1+l+l=r+10 (we have a carry over from the 1's place and need a carry over for the 100's place). Thus, r is odd, so r=7.

g=1: from the 100,000's place, 1+5+g=7, so g=1.

l=8: from the 10's place, 1+l+l=7+10, so l=8.

n=6,b=3: from the 1,000's place, n+7=b+10. so, n=b+3. We only have 2,3, and 6 left, so n=6, b=3.

o=2: it's the only number left

PP

[lorinda]

I messed up, but found an amazing solution anyway!!!!!!!!!

I accidentally missed subtituting one of the 9s for an E!!

So i was looking at

5oNAL5 +
G9RAL5 =
RoBER0

so, now it got interesting, of course I managed to miss my oversight, and continued....shouldnt be possible of course

from there:
L=3 or 8
if it is 3 then A can take ANY remaining value, so I tried L=8,then A can only be 6

giving

5oN685
197685 =
7oB370

and only 2 and 4 as my remaining values.
At this point I cried "Eureka" and came to post my answer to the trick question (letting o = 0) only to find pseudo's post with the correct answer, so my

504685 +
197685 =
702370

is sadly wrong, but it works for
DONALD + GERALD = ROBURT

[Bozeman]

Ok, I will assume that each letter is a unique digit. If I find a solution with this assumption, good, if not, I will relax the assumption.

D=5, therefore T=0, and R>5. Also, L+L+1=R or R+10, so R is odd. From the second digit, and the fact that 0 is already taken, we know that E=1 or 9. If it is 1, then A=5, but T=5, so thus E=9 and A=4 (A+A=E or E-1). Therefore R=7, and R+10=L+L+1, so L=8. The Third digit tells us that N+7=10+B, so N=B+3. The only available choices are N=6, B=3. This leaves G and O. D+G+1=R, so G=1, therefore O=2. (note: no constraint is applied to O by the second digit).

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

and 526485+197485=723970.

Craig

DONALD + GERALD = ROBERT

All letters represent distinct numbers 0-9.

1. Given D=5
2. D + D = T Therefore T = 0
3. D+G <=R by given word
4. R <= 9 by definition
5. 5+G <=R by 1 and 3
6. G <=4 by 4 and 5
7. R > 5 by 1 and 3 and 6
8. R is odd because L + L + 1 = R or 1R by word and 1
9. R is 7 or 9 by 7 and 8
10. L = 3 or 4 or 8 or 9 by 9 and 2L+1 = R or 1R
11. N+R > 10 because O+E=O otherwise and E <> 0 (10,000 column)
12. O + E + = O by 11 and 10,000 column the only way this is true is if E = 9
13. R = 7 by 9 and 12
14. A+A = E or 1E or 2A +1 = E or 1E by 100's
15. 2A = 9 or 19 or 2A + 1 = 9 or 19 by 12 and 14.
16. 2A+1 = 9 or 19 by 15 and defn of odd
17. 2A = 8 or 18; A = 4 or 9 can't be 9 by 12 os A = 4
18. 2L+1 = 7 or 17 by 8 and 13.
19. 2L+1 = 17 because 16 had to have a remainder
20. 2L = 16 by 19 therefore L = 8
21. N + R = B (no remainder for sure because A = 4) by 1000's
22. N + 7 = B
23. N>=4 because otherwise O + E = O can not resolve
24. N = 6 by 23 and the fact that it is the only number left >=4
25. B = 3 by 22 and 24.
26. only numbers remaining are 1 and 2
27. 5 + G + 1= 7 by 100,000's and 12 and the fact that O+E=O has to cause a carry becaue E = 9 and O <> 0
28. G = 1 by 27
29. O = 2 because it is the last number left

Lets see if it makes sense.

526,485 + 197,485 = 723,970

Kind of long and drawn out... I just started going and seen where I would end up.

[DJA]

[Homer]

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