Author 
Message 
Prince

Posted: Thu Feb 12, 2004 10:33 am Post subject: perfect numbers... 


i kno this is easy (mayb its jus my laziness) does sumone kno of a simple way to find perfect numbers?






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Tony





Prince

Posted: Thu Feb 12, 2004 10:40 am Post subject: (No subject) 


a number thats equal to the sum of its proper factors is a perfect number






Tony

Posted: Thu Feb 12, 2004 10:42 am Post subject: (No subject) 


hmm... forloop factors and see if they make up a perfect number

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Prince

Posted: Thu Feb 12, 2004 10:59 am Post subject: (No subject) 


lol umm could u explain that a bit






Tony





wtd

Posted: Thu Feb 12, 2004 3:17 pm Post subject: (No subject) 


http://mathforum.org/dr.math/faq/faq.perfect.html
And quickly, the Quick BASIC program in the above page for finding Mersenne primes and their associated perfect numbers:
code:  DEFDBL AY: DEFSTR Z: CLS
mer = 1: two# = 1: power = 1: xm = y: x2 = y
1 power = power + 1: mer = mer + mer + 1: two# = two# + two#
LOCATE 25, 1: PRINT "Testing"; mer;
FOR i = 3 TO INT(SQR(mer + 1)) STEP 2: IF (mer MOD i) = 0 THEN 1
NEXT
LOCATE 24, 1: PRINT USING "#,###########"; mer;
PRINT " = 2^"; LTRIM$(STR$(power)); "1";
IF power < 31 THEN
PRINT TAB(36); USING "###,###########"; two# * mer;
ELSE PRINT TAB(26); "2,305,843,008,139,952,128";
END IF
PRINT " = (2^"; LTRIM$(STR$(power)); "1) * 2^"; LTRIM$(STR$(power  1))
IF power < 31 THEN 1 






Cervantes

Posted: Thu Feb 12, 2004 5:10 pm Post subject: (No subject) 


um, anyways... google prime numbers you'll find a bunch of stuff.
its something like 2(n^2) (2n^2) where (2n^2) is a prime... dunno, kinda confusing.. google it though, you'll find a much better explanation






MysticVegeta

Posted: Fri May 06, 2005 5:56 pm Post subject: (No subject) 


Tony wrote: well I'm not too sure on the "proper factors" part
Yes they are only proper factors.






