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 13.2 the "Monster" 
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Joined: 07 May 2009, 16:45
Posts: 62
Post 13.2 the "Monster"
Page 254, the order of the largest of the exceptional finite simple groups, the monster, is given as a number divisible by 10^9, that is, it has nine trailing zeros. Seems kind of strange that as humans we have ten fingers, so we have a base 10 number system, so we consider one billion a special number, and this monster happens to fall so evenly on that boundary of a billion. However, if we evolved with 12 fingers, and had a base 12 number system, the monster would end in 20 zeros. (2^{46} \times 3^{20} = 12^{20} \times 2^{26}) Yes? So maybe it is not as strange as it first looks.


17 Aug 2009, 05:05
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Joined: 07 Jun 2008, 08:21
Posts: 235
Post Re: 13.2 the "Monster"
DimBulb wrote:
(2^{46} \times 3^{20} = 12^{20} \times 2^{26}) Yes?

Not quite....

2^{46} \times 3^{20}=(2^2)^{23}\times 3^{20}=(2^2)^{20}\times (2^2)^{3}\times 3^{20}=(2^2)^{20}\times3^{20}\times(2^2)^{3}=12^{20}\times 2^6


24 Aug 2009, 19:28
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