Timeline for The London safes and their secret combinations
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Oct 1, 2015 at 10:44 | comment | added | dr_ | @KBusc If you solve the equations you'll see that this is the only solution. You can find other values that hold true if you consider a smaller subset of numbers (e.g. from 1 to 10 safes). | |
| Oct 1, 2015 at 7:25 | comment | added | Christoph | @SeraphCheng I added some explanation | |
| Oct 1, 2015 at 7:24 | history | edited | Christoph | CC BY-SA 3.0 |
added 1258 characters in body
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| Oct 1, 2015 at 6:07 | comment | added | Saurabh Prajapati | after reading all comments and thinking for two an hour , i still unable to understand the answer. can you please explain more and step by step please. | |
| Sep 30, 2015 at 19:42 | comment | added | KBusc | Any proof that this is the ONLY solution? I'm not convinced this is unique. | |
| Sep 30, 2015 at 18:20 | comment | added | S.C. | Excellent. How did you get to this solution? | |
| Sep 30, 2015 at 15:45 | comment | added | Puzzle Prime | @dr01, oh I see, it seems you have changed it from 36 to 25 since the previous time I checked. | |
| Sep 30, 2015 at 15:09 | comment | added | dr_ | @ArturKirkoryan Unfortunately it's not possible to fit 18 in the pattern: T is supposed to be 0 because of EIGHTEEN=EIGHT+TEN, but it was previously set as 11 by other constraints. | |
| Sep 30, 2015 at 15:03 | comment | added | Puzzle Prime | @dr01, yeah I guessed that's why you multiplied by 2. Still, it was very close for the system to be unsolvable, which is what makes your problem so nice. | |
| Sep 30, 2015 at 14:57 | comment | added | dr_ | @ArturKirkoryan Using the same values works well for numbers up to 10. For bigger numbers, it must be E=0 (because SIXTEEN = SIX+TEN) and I=0 (because THIRTEEN=THREE+TEN). If you're working with the same values, this makes N=4.5 because of NINE=9. The reason I multiplied by 2 was also to have a riddle with only integers. | |
| Sep 30, 2015 at 14:49 | comment | added | Puzzle Prime | @dr01, I guessed your system didn't have a solution for ONE=1, TWO=2, etc. due to problems modulo 2 and this is why you multiplied by 2. Now when I think about it though, 14, 16, 17, 19 repeat the digits 4, 6, 7, 9 with added teen at the end, so basically we have 16 degrees of freedom and 16 numbers to encode (1-13+15+18+20), which generally should be possible. Still, the coincidence is interesting and also a bit baffling why you have so many zeros among the letter values. | |
| Sep 30, 2015 at 14:36 | comment | added | dr_ | @ArturKirkoryan Thanks. It is not mandatory to multiply by 2. This kind of mathematical recreation (known as neomerology) is usually done with the same values i.e. ONE=1, TWO=2, etc. I just wanted to make a riddle that was somehow different. | |
| Sep 30, 2015 at 14:28 | comment | added | Puzzle Prime | @Aioros, after 20 it is not a surprise it works (since "Y" appears for the first time in "twenty" and 0-9 repeat all the time), but it is interesting how with just 16 letters (12 non-zeros) you can get all of the first 20 numbers (I assume this is why the OP had to multiply by 2?). Very nice indeed. | |
| Sep 30, 2015 at 14:22 | comment | added | Aioros | Nah, not for me at least. I definitely wouldn't have figured it out all the same. | |
| Sep 30, 2015 at 14:21 | comment | added | dr_ | @Aioros Thanks. I apologize again for my error; it slipped somehow during the calculations of the Diophantine equations for the 90 codes. It shouldn't have spoiled too much the riddle. | |
| Sep 30, 2015 at 14:15 | comment | added | Aioros | I wanted to say that the error in the original question threw us a bit off and that the number replacements are too weird, but I'm entirely too baffled from seeing that this actually works somehow. | |
| Sep 30, 2015 at 14:05 | comment | added | dr_ | Yes, that was it. Congratulations! | |
| Sep 30, 2015 at 14:04 | history | bounty ended | dr_ | ||
| Sep 30, 2015 at 14:04 | vote | accept | dr_ | ||
| Sep 30, 2015 at 14:03 | history | answered | Christoph | CC BY-SA 3.0 |