Prin merikous mhnes eixa steilei ena quiz me to parapanw subject. Epeidh
kapou yphrxe ena la0os (to opoio to eida meta me th bohtheia toy filou
Kwsta Tampourh), kante update to arxeio sas me th swsth lysh.
To 0ema htan poios o megalyteros arithmos o opoios mporei na ftiaxtei me
ta pshfia 1,2,3 (apo mia fora to ka0ena).
H apanthsh poy eixa dwsei htan
"
[.1^(-32)]! peripoy iso me (2^100)!
(poy einai to mhden reeeeeeeeeeee) ;)
"
Alla h swsth einai {.1^[(-32)!]}
Kai parathetw th dikaiologhsh tou ktamp wste na to deite kai eseis.
Filika
Yiannis
---------- Forwarded message ----------
Date: Tue, 21 Sep 1999 11:13:00 +0300
From: Konstantinos Tampouris <ktamp@cc.uoa.gr>
To: ÃéÜííçò ÊåñêéíÝò <jkerk@arnold.chem.uoa.gr>
Subject: Re: Quiz 1,2,3: Analush
Kai omws epimenw na diafwnw (h Gh gyrizei).
Antimetopizw to problhma metatrepontas ta panta se dunameis toy 2, h
periorizontas ta apo dunameis tou 2. An breis kapou la8os pes to mou.
Kwstas Tampourhs
-------------------------------------------------------------------------
Sugkrinw ta (3^21)!, (2^31)!, 3^(21!) kai 2^(31!)
Metasxhmatismoi:
3^(21!) = 2^(log2(3) * 21!)
(2^31)! = (2^31) * (2^31 - 1) * (2^31 - 2) * ... * 2 * 1 <
< (2^31) * (2^31) * (2^31) * ... * (2^31) * (2^31) =
= (2^31)^(2^31) = 2^(31 * 2^31)
(3^21)! = (3^21) * (3^21 - 1) * (3^21 - 2) * ... * 2 * 1 <
< (3^21)^(3^21) = 3^(21 * 3^21) = 2^(log2(3) * 21 * 3^21)
Ek8etes:
31! = 8222838654177922817725562880000000
log2(3) * 21! = 80977227468672572059.42377613...
31 * 2^31 = 66571993088
log2(3) * 21 * 3^21 = 16579267571.05343587...
Sumperasma:
(3^21)!, (2^31)! < 3^(21!) < 2^(31!)
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