Timeline for The London safes and their secret combinations
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 30, 2015 at 15:45 | history | edited | 2012rcampion | CC BY-SA 3.0 |
added f(7) based on information in a comment to the question
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| Sep 30, 2015 at 6:58 | comment | added | dr_ | @DrXorile The combination code of safe $n$ is a function of safe $n$ only. | |
| Sep 29, 2015 at 17:20 | comment | added | 2012rcampion | @DrXorile $C(n)$ is not a function of $C(n-1)$. | |
| Sep 29, 2015 at 13:38 | comment | added | Dr Xorile | Is the formula a function only of the safe, or also of the code from the previous safe? | |
| Sep 29, 2015 at 7:20 | comment | added | dr_ | @2012rcampion Yes, that's an effect of the pattern. | |
| Sep 28, 2015 at 22:57 | comment | added | 2012rcampion | @dr01 Let my try asking in a different way: if the numbers of two safes have the same tens digit, will the difference in their combinations always be twice the difference in their numbers? | |
| Sep 28, 2015 at 8:06 | comment | added | dr_ | @2012rcampion In this sense, yes, there is a function that can be applied to get the combination of any safe -- otherwise there would be no riddle to solve :) | |
| Sep 27, 2015 at 20:48 | comment | added | 2012rcampion | @dr01 Remember that a mathematical function is any mapping from one set to another... so the answer must be a function (if the question is well-posed). I might restate my question as, "is there some choice of $f$ such that the above $C$ produces the same results as your procedure for choosing the combination?" | |
| Sep 27, 2015 at 19:57 | comment | added | dr_ | @2012rcampion Yes, the secret pattern can't be found by a mathematical function alone. (Or, at least, that would surprise me!) Your post contains the solution; take some time to develop it. Tomorrow I'll post another hint if nobody has made any progress. | |
| Sep 27, 2015 at 17:01 | comment | added | 2012rcampion | @dr01 Is that a hint that my $C(n)$ is wrong? | |
| Sep 26, 2015 at 8:24 | comment | added | dr_ | +1 for getting close to the solution! Hint: the solution doesn't involve complex mathematical formulas. | |
| Sep 25, 2015 at 20:39 | comment | added | 2012rcampion | @f'' Good catch! I actually never noticed the missing "u" before! | |
| Sep 25, 2015 at 20:37 | comment | added | f'' | Forty is spelled differently from fourteen, maybe. | |
| Sep 25, 2015 at 20:21 | history | answered | 2012rcampion | CC BY-SA 3.0 |