## across 3 years ago This one should be fun: http://dl.dropbox.com/u/12189012/fact.html

1. SUROJ

2 137 577 22689565741

2. Savvy

it's prime.... ;)

3. across

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

4. Savvy

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

5. SUROJ

I use some algorithms online

6. across

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

7. across

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

8. Savvy

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

9. SUROJ

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

10. across

This number has only two factors.

11. SUROJ

yes.....got it

12. Savvy

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

13. across

I already said that the number is not prime!

14. Savvy

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

15. across

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

16. across

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

17. Savvy

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

18. Zarkon

2158122193002952449690243008233157266924190689, 24058327474941875553768074062128402939746847713

19. across

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

20. across

Zarkon, how did you do that? ;P

21. across

I wonder if you wrote a program.

22. Zarkon

I'm part machine.

23. apoorvk

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

24. sasogeek

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

25. FoolForMath

Not just any program, we need a very fast prime factorization algorithm is required. If you are very confident about your algorithm , you may consider testing it here:www.spoj.pl/problems/FACT2/