tag:blogger.com,1999:blog-32999748.post4546849741938819440..comments2024-03-04T01:47:18.750-05:00Comments on Lemurian Congress: Donald, Gerald, and RobertAdam Thorntonhttp://www.blogger.com/profile/05634565262440008573noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-32999748.post-13263297522127247852009-04-08T13:10:00.000-04:002009-04-08T13:10:00.000-04:00Good Eric!I have just read the puzzle (to take hin...Good Eric!<BR/>I have just read the puzzle (to take hint :-))and it's interesting because a grandfather has sent it to me!!Polohttps://www.blogger.com/profile/16370518563746732881noreply@blogger.comtag:blogger.com,1999:blog-32999748.post-66389519041377360762008-06-11T21:32:00.000-04:002008-06-11T21:32:00.000-04:00Solve: D O N A L D + G E R A L D = R O B E R T, ...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.<BR/><BR/>_<BR/>_ D O N A L D<BR/>+ G E R A L D<BR/>_ == == == == == ==<BR/>_ R O B E R T<BR/> <BR/> <BR/>1. Since D = 5, and D + D = T, then 5 + 5 = 10, so T is 0, and we carry a 1:<BR/>_ _ _ _ _ _ 1<BR/>_ 5 O N A L 5<BR/>+ G E R A L 5<BR/>_ == == == == == ==<BR/>_ R O B E R 0<BR/> <BR/> <BR/>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). <BR/>We observe that O + 9 = O, which means that the result is > 10, and we must have a carry-over of 1. We have:<BR/>_ _ 1 _ _ _ 1<BR/>_ 5 O N A L 5<BR/>+ G 9 R A L 5<BR/>_ == == == == == ==<BR/>_ R O B 9 R 0<BR/><BR/>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.<BR/>_ _ 1 _ _ 1 1<BR/>_ 5 O N 4 L 5<BR/>+ G 9 R 4 L 5<BR/>_ == == == == == ==<BR/>_ R O B 9 R 0<BR/><BR/>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)<BR/><BR/>Digits used up: { 0, 4, 5, 9 } Free: { 1, 2, 3, 6, 7, 8 }<BR/>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}. <BR/>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. <BR/><BR/>We now try to figure out if R is 3 or 7.<BR/>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). <BR/><BR/>Thus, R must take value 7. We have:<BR/> <BR/><BR/>_ _ 1 1 _ 1 1<BR/>_ 5 O N 4 L 5<BR/>+ G 9 7 4 L 5<BR/>_ == == == == == ==<BR/>_ 7 O B 9 7 0<BR/><BR/>5. Since 5 + G + 1 = 7, then G = 1. We have:<BR/>_ _ 1 1 _ 1 1<BR/>_ 5 O N 4 L 5<BR/>+ 1 9 7 4 L 5<BR/>_ == == == == == ==<BR/>_ 7 O B 9 7 0<BR/><BR/>6. Since L + L + 1 = 7, then we must have L = 3 or L = 8. <BR/>We notice from 4 + 4 = 9 that we must be carrying a 1 over from L + L.<BR/>Since we carry a 1 from L + L, then we cannot have L = 3, and therefore L = 8:<BR/>_ _ 1 1 _ 1 1<BR/>_ 5 O N 4 8 5<BR/>+ 1 9 7 4 8 5<BR/>_ == == == == == ==<BR/>_ 7 O B 9 7 0<BR/><BR/>7. Digits used up: {1, 4, 5, 7, 8, 9, 0} Free: {2, 3, 6}<BR/>We try to see which free digits match next.<BR/>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.<BR/>It immediately follows that B = 3 (since 6 + 7 = 13)<BR/>_ _ 1 1 _ 1 1<BR/>_ 5 O 6 4 8 5<BR/>+ 1 9 7 4 8 5<BR/>_ == == == == == ==<BR/>_ 7 O 3 9 7 0<BR/><BR/>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:<BR/>_ _ 1 1 _ 1 1<BR/>_ 5 2 6 4 8 5<BR/>+ 1 9 7 4 8 5<BR/>_ == == == == == ==<BR/>_ 7 2 3 9 7 0<BR/><BR/>Adding 526485 with197485 gives the result: 723970, therefore the addition is correct. The mapping is thus:<BR/>0 = T, 1 = G, 2 = O, 3 = B, 4 = A, 5 = D, 6 = N, 7 = R, 8 = L, 9 = E. <BR/><BR/>(2008, by Alan Lupsha)Alanhttps://www.blogger.com/profile/04661881652565121167noreply@blogger.comtag:blogger.com,1999:blog-32999748.post-78993890727656907972007-06-30T11:23:00.000-04:002007-06-30T11:23:00.000-04:00Pretty good, Eric!Yeah, have to admit, math puzzle...Pretty good, Eric!<BR/><BR/>Yeah, have to admit, math puzzles make my spine shrink.<BR/><BR/>Hats off to you, though, Muffy, for mastering them as much as you have!Harryhttps://www.blogger.com/profile/02731815508918636273noreply@blogger.comtag:blogger.com,1999:blog-32999748.post-853301683167218692007-06-29T16:58:00.000-04:002007-06-29T16:58:00.000-04:00Sorry--that side of my brain has had a "For Rent" ...Sorry--that side of my brain has had a "For Rent" sign up for years.<BR/><BR/>I look at those names and think, "Duck + Ford = Frost."Eric Littlehttps://www.blogger.com/profile/02556454801310628473noreply@blogger.com