Luke_A_Somers comments on The Ultimate Newcomb's Problem - Less Wrong

18 Post author: Eliezer_Yudkowsky 10 September 2013 02:03AM

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Comment author: Luke_A_Somers 11 September 2013 08:13:48PM *  -1 points [-]

Primes less than sqrt(1033) for which I know of no really obvious tricks (i.e. the digit adding tricks for them aren't so simple one can trivially do them in your head): 7, 13, 17, 19, 23, 29, 31

Also, since the scenario said we've done this many times and we haven't been trolleyed yet, it can't be all that easy to get trolleyed.

(eta: why the -1? Both points seem solid to me, even in the light of the additional trick below - there are several possible factors remaining, and it's not like I was enumerating these during my 2 minutes. Moreover, the 1001 trick works for 1033, but doesn't help so much with other typical numbers of that general magnitude - say, 1537, that it's something you're liable to do by accident)

Comment author: [deleted] 14 September 2013 02:21:02PM 1 point [-]

Also, since the scenario said we've done this many times and we haven't been trolleyed yet, it can't be all that easy to get trolleyed.

Maybe you get run over by the trolley only if you try to factor Omega's number, and all the times before this Omega's number and the lottery number were different, and there's no much point in trying to factor the lottery number in that case since (assuming linear utility of money) it has no relevance on how many boxes you should take.

Comment author: [deleted] 12 September 2013 08:18:21AM *  0 points [-]

1001 = 7 x 11 x 13, so each of those divides 1033 iff it divides 32 (and divides 314,159,265 iff it divides 265 - 159 + 314).