perfect numbers...

[Prince]

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

[Tony]

what's the definition of a perfect number?

[Prince]

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

[Tony]

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

[Prince]

lol umm could u explain that a bit

[Tony]

well I'm not too sure on the "proper factors" part... but assuming that any number can be a proper factor and you'll looking at numbers up to 100 assuming each has 2 factors only.

code:

for(i = 1; i<11;i++) {      for(i2 = 1; i2<11;i2++)      {           if (i*i2) == (i+i2) then           { System.out.println(i+i2 + " is a perfect number); }      } }

Thought I donno... it seems that there got to be a better way out there...

[wtd]

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 A-Y: 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]

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]

Tony wrote:

well I'm not too sure on the "proper factors" part

Yes they are only proper factors.