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SUROJ
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2
137
577
22689565741
Savvy
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it's prime.... ;)
across
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Where did you get those numbers from? And it's not a prime number.
Savvy
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are you actually serious about solving that....????
SUROJ
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I use some algorithms online
across
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I guess you're right: this website is not a serious place. :)
across
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Your algorithms are wrong because the number ends in 7, yet you say that it's divisible by 2. :P
Savvy
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actually this is a nyc place but that no. is a bit too large to solve...
SUROJ
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oh yea......so it should be only itself...
across
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2
This number has only two factors.
SUROJ
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yes.....got it
Savvy
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then they have to be 1 and itself and hence its prime...
across
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I already said that the number is not prime!
Savvy
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then how can it have ONLY 2 factors.....1 and itself have to be compulsory factors....
across
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With two factors, I obviously meant to state proper ones (and excluding 1).
across
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Otherwise, the problem would've been trivial, no? ;P
Savvy
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with such a LARGE no. the problem can never be a trivial one.....
Zarkon
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1
2158122193002952449690243008233157266924190689,
24058327474941875553768074062128402939746847713
across
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I said that the problem would have been trivial if it had been a prime.
across
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Zarkon, how did you do that? ;P
across
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I wonder if you wrote a program.
Zarkon
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I'm part machine.
apoorvk
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so, how do we 'solve' this? o.O
sasogeek
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i'm going to write a program to solve this :) and yes, should be fun :D
FoolForMath
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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/