rookie1ja Posted March 31, 2007 Report Share Posted March 31, 2007 No. 3 - Back to the Cryptograms and Algebra Puzzles Replace the same characters by the same numerals so that the mathematical operations are correct. RE + MI = FA DO + SI = MI LA + SI = SOL This old topic is locked since solution is already provided in the Spoiler below. Pls visit New Puzzles section to see always fresh brain teasers. No. 3 - solution 27 + 56 = 83 40 + 16 = 56 93 + 16 = 109 Link to comment Share on other sites More sharing options...
Guest Posted June 9, 2007 Report Share Posted June 9, 2007 Could the answer also be the following ... 1+0 = 1 (RE+MI = FA) 0+0 = 0 (DO+SI = MI) 1+0 = 1 (LA+SI = SOL) Enjoy the site, came across while solving "Who owns the Fish"!!! Link to comment Share on other sites More sharing options...
Guest Posted September 26, 2007 Report Share Posted September 26, 2007 Replace the same characters by the same numerals so that the mathematical operations are correct. RE + MI = FA DO + SI = MI LA + SI = SOL LA + SI = SOL implies S = 1. LA + 1I = 1OL would cause L to be 8 or 9, but in both cases, O would have to be 0, which also is in line with DO + SI = MI A and I can't both be high enough to get L = 8 and carry over into O, so L must be 9, thus 9A + 1I = 109. A + I must thus be 9 From DO + 1I = MI then follows M = D + 1. So far we have: RE + MI = FA D0 + 1I = MI 9A + 1I = 109 A can not be 0, 1, or 9, which are already in use. A can not be 8, because I would be 1 (used). A can not be 7, because then I would be 2, and E 5, but that would leave 3, 4, 6 and 8 which could in no way fit D, F, M and R. A can not be 6, because both I and E would then be 3. A can not be 5, because that would make I = 4, and thus E is 1 (used). A can not be 4, because I would become 5, and thus E would be 9 (used). Assume A = 3, this would imply I = 6, E = 7. Then 2, 4, 5 and 8 would remain. R = 2, D = 4, M = 5, F = 8 will fit. Assume A = 2, this would imply I = 7, E = 5. Then 3, 5, 6, and 8 would remain. These can not be put for D, F, M and R in any mathematically correct fashion. So the correct solution would be S = 1, R = 2, A = 3, D = 4, M = 5, I = 6, E = 7, F = 8, L = 9, O = 0 27 + 56 = 83 40 + 16 = 56 93 + 16 = 109 Thanks again, BoilingOil Link to comment Share on other sites More sharing options...
Guest Posted February 24, 2008 Report Share Posted February 24, 2008 I don't know if this solution is correct, but I've found: FA=9 RE=4 MI=5 DO=3 SI=2 LA=6 SOL=8 So: RE+MI=FA 4+5=9 SI+DO=MI = 2+3=5 SI+LA=SOL 2+6=8 It's an easy solution. I've found it by deduction, but I'm not sure that I didn't forget to read some condition for the problem (my english isn't very good... to be honest, it's very bad ). If someone has been able to read my strange english, please, say me if I'm right! Bye! (It's my first message. I've logged in just now... ^.^) Link to comment Share on other sites More sharing options...
rookie1ja Posted February 24, 2008 Author Report Share Posted February 24, 2008 If someone has been able to read my strange english, please, say me if I'm right! 1 letter shall be 1 unique number ... so, for instance, FA can not be just 1 number (9) - it has to be 2 numbers (1 for F and 1 for A) Link to comment Share on other sites More sharing options...
Guest Posted December 10, 2010 Report Share Posted December 10, 2010 There is another solution possible assuming that one of the alphabets can be a two digit number (nothing in the question suggests that this is not possible) O=0 I=2 S=1 L=9 A=7 E=5 M=4 D=3 R=6 F=10 Now we have, 1. RE+MI=FA 65+42=107 2. DO +SI=MI 30+12=42 3. LA + SI=SOL 97+12=109 Link to comment Share on other sites More sharing options...
Guest Posted December 10, 2010 Report Share Posted December 10, 2010 In the above case, R=8 and F=12 is also a possible solution(the values for the other alphabets are the same) Link to comment Share on other sites More sharing options...
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