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 Add the process you went trought whit the answer. |
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 |
that was my post *NM*
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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. |
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 |
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 |
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>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 |
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 <=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. |
Above Post is Mine
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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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