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2
137
577
22689565741

it's prime.... ;)

Where did you get those numbers from? And it's not a prime number.

are you actually serious about solving that....????

I use some algorithms online

I guess you're right: this website is not a serious place. :)

Your algorithms are wrong because the number ends in 7, yet you say that it's divisible by 2. :P

actually this is a nyc place but that no. is a bit too large to solve...

oh yea......so it should be only itself...

This number has only two factors.

yes.....got it

then they have to be 1 and itself and hence its prime...

I already said that the number is not prime!

then how can it have ONLY 2 factors.....1 and itself have to be compulsory factors....

With two factors, I obviously meant to state proper ones (and excluding 1).

Otherwise, the problem would've been trivial, no? ;P

with such a LARGE no. the problem can never be a trivial one.....

2158122193002952449690243008233157266924190689,
24058327474941875553768074062128402939746847713

I said that the problem would have been trivial if it had been a prime.

Zarkon, how did you do that? ;P

I wonder if you wrote a program.

I'm part machine.

so, how do we 'solve' this? o.O

i'm going to write a program to solve this :) and yes, should be fun :D