How do you get factor pairs of big numbers? - Answers

Well, for a given value of big . . .If you have a really big number which is the product of two primes (maybe 100 digits each) you are looking for the current Holy Grail of some workers in number theory.Generating numbers around 100 digits long is dead easy. Identifying the primes amongst them takes longer, but is quite do-able. Multiplying two of these primes to make a 200 digit composite number is, again, straightforward.Now try undoing it. Given a 200 digit number that was produced by multiplying two primes, find those primes. As far as we know, no-one has come up with a way of doing this within a lifetime of computation, even using the fastest number cruncher around.There are encryption systems in use that rely on the practical impossibility of factorising these huge numbers; if you can find a way to do it you could probably write your own check at GCHQ or the NSA.The same as any other number. It will probably take a little longer since there's more factors. First step: Find the prime factorization.23 x 32 x 52 x 11 = 19800 That gives you 2, 3, 5 and 11. Every other factor is in there in some combination. The rules of divisibility give you 4, 6, 8, 9 and 10. That gives you 10 factor pairs. (19800,1)(9900,2)(6600,3)(4950,4)(3960,5)(3300,6)(2475,8)(2200,9)(1980,10)(1800,11) Work your way up from there. 22 x 3, 3 x 5, 2 x 32, 22 x 5, 2 x 11, 23 x 3, 5 x 5, 2 x 3 x 5, 3 x 11, 22 x 32 (1650,12)(1320,15)(1100,18)(990,20)(900,22)(825,24)(792,25)(660,30)(600,33)(550,36) There are sixteen more.