Finding all divisors of a number

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To find all divisors of a number, it is sufficient to divide only up to its square root rather than its half. Prime factorization simplifies the process, as identifying prime factors allows for the determination of all divisors. For large numbers, advanced methods like Pollard's rho algorithm and elliptic curve factoring can expedite the factoring process. The discussion highlights the importance of recognizing that factors greater than the square root can still be identified through their corresponding smaller factors. Overall, understanding these techniques can significantly enhance efficiency in divisor finding.
ShayanJ
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Good time people
Is there any way other than dividing the number from 2 to its half to find all of its divisors?
Anyway may help so feel free telling your ideas
thanks
 
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First of all you don't have to divide up to the number's half, only up to it square root. Second, you only have to divide by prime numbers.
 
It's easiest to first find all the prime factors of a number, and then determine the divisors from that list. Finding the prime factors requires only trial dividing up to the square root of the number -- and then only by the primes, if you like.

Methods like Pollard's rho algorithm, elliptic curve factoring, and the number field sieves can be used to factor large numbers (say, > 10^9) much more quickly.

Practically, if you don't recognize these terms, I recommend a tool like
http://www.alpertron.com.ar/ECM.HTM
 
Hello
Did you noticed 71,a prime factor of 284?
it is greater than 16.85 that is the square root of 284.
 
And one more thing...
Could you explain the methods you named?
the page you gave the link of,doesn't include a explanation about the method itself.
I searched it in the internet but i found no easy explanation.
thanks
 
Hi Shyan,

I had visited a site called eTutorWorld its online tuition site.
 
Shyan said:
Hello
Did you noticed 71,a prime factor of 284?
it is greater than 16.85 that is the square root of 284.

Yes, but 284=71*4. So by finding 4, you've found 71 as a factor also. If a> sqrt(n), and ab = n, then b<sqrt(n) necessarily.
 
Yes, but I think Shyan was referring to Skeptic2's advice to only go up to and only use prime numbers. 2 divides into 284 142 times but you don't immediately see the "71". Of course the intelligent thing to do once you have 284= 142 would be to ignore the 284 and start factoring 142 so that you immediately get 2*71 but it is not clear that was what Skeptic2 meant.
 
HallsofIvy said:
Yes, but I think Shyan was referring to Skeptic2's advice to only go up to and only use prime numbers. 2 divides into 284 142 times but you don't immediately see the "71". Of course the intelligent thing to do once you have 284= 142 would be to ignore the 284 and start factoring 142 so that you immediately get 2*71 but it is not clear that was what Skeptic2 meant.

Agreed.
 
  • #10
shyan said:
Could you explain the methods you named?
Please!
 

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