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#1
10-15-2002, 01:20 AM
 happyjaypee Senior Member Join Date: Sep 2002 Location: Quebec, Canada Posts: 517
Bartlett\'s problem

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

#2
10-15-2002, 04:18 AM
 Guest Posts: n/a
Re: Bartlett\'s problem

2 more things:
All letters have a value of 0 to 9
Two differents lettres can't have the same value
#3
10-15-2002, 04:20 AM
 happyjaypee Senior Member Join Date: Sep 2002 Location: Quebec, Canada Posts: 517
that was my post *NM*

#4
10-15-2002, 06:21 PM
 lorinda Senior Member Join Date: Sep 2002 Location: England Posts: 2,478
Re: Bartlett\'s problem

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.
#5
10-15-2002, 06:54 PM
 PseudoPserious Senior Member Join Date: Oct 2002 Posts: 151
I\'ll take a stab...

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
#6
10-15-2002, 07:27 PM
 lorinda Senior Member Join Date: Sep 2002 Location: England Posts: 2,478
Re: I\'ll take a stab...

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
#7
10-15-2002, 08:14 PM
 Bozeman Senior Member Join Date: Sep 2002 Location: On the road again Posts: 1,213
Re: Bartlett\'s problem

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&gt;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
#8
10-16-2002, 01:16 AM
 Guest Posts: n/a
Re: Bartlett\'s problem

DONALD + GERALD = ROBERT

All letters represent distinct numbers 0-9.

1. Given D=5
2. D + D = T Therefore T = 0
3. D+G &lt;=R by given word
4. R &lt;= 9 by definition
5. 5+G &lt;=R by 1 and 3
6. G &lt;=4 by 4 and 5
7. R &gt; 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 &gt; 10 because O+E=O otherwise and E &lt;&gt; 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&gt;=4 because otherwise O + E = O can not resolve
24. N = 6 by 23 and the fact that it is the only number left &gt;=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 &lt;&gt; 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.

#9
10-16-2002, 01:18 AM
 DJA Senior Member Join Date: Sep 2002 Posts: 209
Above Post is Mine

#10
10-16-2002, 12:20 PM
 Homer Senior Member Join Date: Sep 2002 Posts: 5,909
Re: Bartlett\'s problem

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

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 &gt;= 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 &gt; 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 &gt; 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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