Dork news: New Mersenne prime discovered

[hurleyfor3]

(2^43112609) - 1 is prime. Last digit has to be... 1.

[2535Miles]

(2^43112609) - 1 is prime. Last digit has to be... 1.

Dang, that's sweet. Finding a 13 million digit prime number ONLY makes you eligible for the $100,000 prize.

[weezie]

Numbers dorks are so cute. Really, smart guys rule!

[YmoBeThere]

The term dorks more often connotes only average intelligence combined with a lack of social grace, etc. Nerds and geeks on the other hand...

[YmoBeThere]

Wow, YBT, that was a dorky statement...

[DukeUsul]

[Lavabe]

Wow, YBT, that was a dorky statement...

"There are children on the streets...
Killing each other using knives and forks
And calling each other names like Dork."

Think about it!
Cheers,
Lavabe

[hurleyfor3]

Wow, YBT, that was a dorky statement...

but not quite as dorky as replying to your own posts

[2535Miles]

but not quite as dorky as replying to your own posts

How is that dorky?

[2535Miles]

How is that dorky?

Yeah, for reals.

[hc5duke]

(2^43112609) - 1... Last digit has to be... 1.

I was going to say, that's not true, but then I realized you can figure out the last digit without knowing the number... and yes, that last digit is 1, Mr. smarty-pants.

[g_olaf]

I was going to say, that's not true, but then I realized you can figure out the last digit without knowing the number... and yes, that last digit is 1, Mr. smarty-pants.

That's kind'a cool... also since for n>2, n will always be odd, number can only end in 1 or 7. Never noticed that before...

[hc5duke]

That's kind'a cool... also since for n>2, n will always be odd, number can only end in 1 or 7. Never noticed that before...

The one's digit in (2^x)-1 can be 1, 3, 5, or 7 (but never 9):

My explanation:
2^1 = 2 --> (2^x)-1 ends in 1 if (x mod 4) is 1
2^2 = 4 --> (2^x)-1 ends in 3 if (x mod 4) is 2
2^3 = 8 --> (2^x)-1 ends in 7 if (x mod 4) is 3
2^4 = 16 --> (2^x)-1 ends in 5 if (x mod 4) is 0
2^5 = 32
2^6 = 64

... (the least significant digit will always cycle 2-4-8-6)

2^43112609 --> (43112609 mod 4) is 1, therefore the number ends in 2 (and number-1 ends in 1)

not a good proof (actually not a proof at all), but I think that's a pretty clear explanation.

[pamtar]

Thanks for sharing Hurley. Really neat stuff. Are there any notorious equations out there still unsolved? How about one only solved by its creator?

[YmoBeThere]

Thanks for sharing Hurley. Really neat stuff. Are there any notorious equations out there still unsolved? How about one only solved by its creator?

Life itself?

[hc5duke]

Life itself?

42. Next question?

[YmoBeThere]

but not quite as dorky as replying to your own posts

I find that on some of the boards that is the best way to have a reasonable conversation!

[hurleyfor3]

Thanks for sharing Hurley. Really neat stuff. Are there any notorious equations out there still unsolved? How about one only solved by its creator?

Among "easy" conjectures, that is, those which require knowing only basic arithmetic and geometry to understand, I think we still have the Goldbach conjecture, the twin prime conjecture, whether there are infinitely many perfect numbers, the 196 problem and the 3x + 1 problem. There aren't a whole lot of simply-stated ones left... who gives a crap about whether P = NP.

Pierre de Fermat implied he had a solution to his own Last Theorem, but he probably didn't. The actual proof used systems of mathematics far more advanced than anything around in Fermat's time.

[hurleyfor3]

The one's digit in (2^x)-1 can be 1, 3, 5, or 7 (but never 9):

As it turns out, other than the first Mersenne prime which is (2^2) - 1 = 3, the power to which 2 is raised must always be odd. So all Mersenne primes end in either 1 or 7.

[hc5duke]

As it turns out, other than the first Mersenne prime which is (2^2) - 1 = 3, the power to which 2 is raised must always be odd. So all Mersenne primes end in either 1 or 7.

Ah, I see what you and g_olaf mean, I was referring to 2^x-1 in general, not Mersenne primes.