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lorcar
11-16-2004, 10:38 PM
how to answer in the quickest way???

danielman
11-17-2004, 04:48 AM
how to answer in the quickest way???

You just have to obtain the prime factor decomposition of the number and then to combine those factors in all possible forms.

If they only ask you only by the NUMBER of divisors, that's a lot easier. It's only a problem of counting. For example:

756 = (3^3) * (2^2) * 7

Any factor you may have will be a combination (product) of these factors. You have available 3 "3"s, 2 "2"s and 1 "7", so

# of ways factor "3" may appear = 3 + 1 = 4 (3^0, 3^1, 3^2, 3^3)
# of ways factor "2" may appear = 2 + 1 = 3 (2^0, 2^1, 2^2)
# of ways factor "7" may appear = 1 + 1 = 2 (7^0, 7^1)

thus: # of ways these factors may be combined = 4*3*2 = 24
=> there are 24 factors

general rule: just multiply the exponents increased by 1

N = (p1^e1)*(p2^e2)....(pn^en) (p1...pn different prime numbers)
#_factors(N) = (e1 + 1)*(e2 + 1)....(en + 1)

sabna
11-18-2004, 12:06 PM
[QUOTE=danielman]You just have to obtain the prime factor decomposition of the number and then to combine those factors in all possible forms.


can u explain please what to do in the above quoted state?..
anyway , a lot of thanks for ur valuable info..!!

prayers
sabna

thebullfighter
11-18-2004, 12:34 PM
go thro. th following threads. th concept would be clear.


http://www.urch.com/forums/showthread.php?t=14463 (for # of factors)

#13 (http://TestMagic.com/forums/showpost.php?p=78714&postcount=13) (for power of prime factor in n! , ps. read the alternate method.)
also, refer to-- http://www.urch.com/forums/showthread.php?t=14462
http://www.urch.com/forums/showthread.php?t=15237