# Fastest way to find all the prime factors of a very large number without calculator? [closed]

1. Largest possible factor of a very large number n would be number itself.
2. The largest would be n/2 (if prime) if n/2 is not prime then it would be less than n/2 .
3. The smallest factor would be 1.

Is there any general approach to find all the remaining prime factors using these three numbers?

Thanks..

## closed as off-topic by RRL, Did, José Carlos Santos, Peter, Trevor GunnJan 28 at 16:44

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• No, there is not. – Randall Jan 22 at 2:57
• Point 1 seems incorrect. The number will not be a prime factor of itself if it is not prime (for example $2^200$ is not prime and so not a prime factor of itself, same goes for smaller numbers like 10). – The Long Night Jan 22 at 3:05
• There is a fastest way to find all the prime factors of a number but I wouldn't call it fast. – Gnumbertester Jan 22 at 3:09
• There isn't even a fast way to write down a very large number, much less find its prime factors. And for very large numbers, even calculators are useless – where will you find a calculator that keeps as many as 100 digits? Computers and exceedingly clever algorithms are the way to go. – Gerry Myerson Jan 22 at 3:17
• A very large number can only be completely factored without a calculator in exceptional cases. The trial division method is far too slow, for a $200$ digit number, lets say, even the fastest computer would be overwhelmed. There are much better algorithms, which you could of course, do by hand as well in principle, but they are more complicated and still would in general need too many steps. If you do not even allow a table calculator (which I assume because of the title) , it will in general be even difficult to factor, lets say, a $9$ digit number.completely. – Peter Jan 22 at 10:22

As you find factors you can divide them out. If there are no factors smaller than $$\sqrt n$$ the number is prime, which is much smaller (for large $$n$$) than $$n/2$$. Factoring is believed to be hard-this is the basis of the security of RSA encryption. It is easy to prove a number composite and relatively easy to prove a number prime, but if you have a large number that is the product of two large primes it is believed to be impractical to find the factors. We don't have a proof that it is hard, but lots of people have tried and failed. If somebody did find a solution to factoring it is not clear they would publicize it because they could use it to decrypt things we believe to be secure.
• To the proposer. $n$ is composite iff $n$ has a $prime$ divisor $p$ such that $p\le \sqrt n.$ – DanielWainfleet Jan 22 at 3:33