# New Largest Known Prime Number – Numberphile

There is a new “largest known prime number”.

Extra footage: http://youtu.be/o0ZOs7sMS7k

More on Mersenne Primes: http://www.youtube.com/watch?v=PLL0mo5rHhk

Perfect Numbers: http://www.youtube.com/watch?v=ZfKTD5lvToE

Googolplex: http://www.youtube.com/watch?v=8GEebx72-qs

Graham’s Number: http://www.youtube.com/watch?v=XTeJ64KD5cg

This video features Dr Tony Padilla from the University of Nottingham.

Website: http://www.numberphile.com/

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Videos by Brady Haran

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Ross Jennings• Oct 15, 2016 #You could have predicted the last digit, you know. The powers of two in base ten end in 2, 4, 8, 6, 2, 4, 8, 6, etc., and 57,885,161 is an even number of tens plus one, so 1 modulo 4, which means the last digit of 2^57,885,161 is 2, and therefore the last digit of 2^57,885,161-1 must be 1.

tenacious645• Oct 15, 2016 #2 ^ (2 ^ 57,885,161 – 1) – 1…..why wouldn't that be a new prime number?

Stewartsaurus• Oct 15, 2016 #The largest number is the number of numbers.

Tune Lah mitunlay• Oct 15, 2016 #I SWEAR RSA SYSTEM OF SECURE IS DOWN ….

Tune Lah mitunlay• Oct 15, 2016 #THE FIRST APPLICATION OF P=NP WAS : WHEN I ASK MYSELF IF A SONG IS IN A RANGE CALLED X AND THE BAD SONG IS ON THE RANGE Y CAN I CALCULATE THE PROBABILITY TO HAVE THE SONG AND PLAY IT ON A KEYBORD .. THE ANSWER IS YES <. BECAUSE I HAD STUK THE SONG ON X ; THAT IS TO SAY I HAD GIVEN MYSELF THE SONG …SO X=1 AND Y=0 ; BUT CAREFILL PROBA 1 MEAN THAT THERE ARE SOME KIND OF NOTE WICH PARTICIPATE FOR THE SONG JUST THE CARDINAL OF THE VIBRATION BUT NOT VERY GOOD SONG ; THAT IS JUST AN APROXIMATION TO GO ARROUND THE TRUE SONG…SO 1=111111111…..1111 OR SOMETHING LIKE 1=000000000…….00000.. YOU SEE THE APPLICATION THAT MAKE PROBABILITY A SURELY…OR AXIOMS DEMONSTRABLE OR IMPOSSIBLE POSSIBLE ?

Tito Titoburg• Oct 15, 2016 #Does that mean that 2^(2^57,885,161-1)-1 is a prime? If it is I just found a larger prime. Throw money at me.

Andy Vo• Oct 15, 2016 #Another nice proof is knowing that all powers of 2 are expressible in binary as a 1 followed by a number of 0s. Eg. 64 in binary is 1000000. It's then easy to see that if you subtract 1 from any of these power of 2 you're just going to get a series of 1s in binary. Eg. 63 is 111111

Audible Magician• Oct 15, 2016 #w 0 w 2^74207281-1 is prime

314rft• Oct 15, 2016 #This video is officially dated as of January 19th, 2016. New prime found: 2^74,207,281 -1.

LycanDragon• Oct 15, 2016 #takes a red pen and crosses out the number in the intro and replaces it with 2^74207281-1Audible Magician• Oct 15, 2016 #I'm just letting you know

31415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066400000069 is a prime

Audible Magician• Oct 15, 2016 #What if someone found a trillion digit prime

WHO KNOWS MAYBE

999999999999999999999^9999999999999999999999999999999999999999

COULD BE PRIME

pe1dnn• Oct 15, 2016 #The new number 2^74207281 – 1 is from Curtis Cooper like the previous 2^57885161 – 1 was. Since it is a Mersenne prime this means also a new perfect number is found!

Kathie Dart• Oct 15, 2016 #(now out of date)

apoorv gupta• Oct 15, 2016 #i calculated biggest prime number with more than 100 millions digit. SHYAM SUNDER GUPTA INDIA

ben1996123• Oct 15, 2016 #w 0 w 2^74207281-1 is prime

Crazy drummer• Oct 15, 2016 #Not anymore

TurboCMinusMinus• Oct 15, 2016 #It would take more than the age of the universe (for a single modern computer, presumably) to go through all of the divisors? So… how did you prove this was prime then? Undoubtedly some fancy-schmancy mathematical test but, that seems a little less than airtight, to me.

paulspydar• Oct 15, 2016 #this all blows my mind, Its like understanding the words but not the language..

Isaac Billingham• Oct 15, 2016 #Tony didn't need to look at the computer screen to work out the last digit!

Here's one way to work it out…

Since p = 2^(57,885,161)-1. We know that 57,885,161 is of the form 4N+1 (where N is an integer) because 57,885,100 is a multiple of 4 and 61 =4*15+1.

We also know that 2^(4N+1) will end in a 2 (look at the sequence 2^N to see this pattern).

Thus 2^(57,885,161) ends in a 2 so p = 2^(57,885,161)-1 must end with a 1.

This kind of argument uses divisibility rules, which turn out to be a very elegant way (in my opinion) of checking little facts about numbers without actually writing them out !

:)

cha rel• Oct 15, 2016 #if we would call this prime number "m", wouldn be 2^m -1 be an even higher prime?

Frank Harr• Oct 15, 2016 #I know this is just a rehash of a sieve, but it's on my mind.

You take a number line. You start at two and eliminate every other number greater than two.

Then you move on to three. You eliminate ever third number greater than that, but not the even ones because they're already gone.

Four is already gone. So you move on to five and get rid of all the remaining multiples of five.

Six is gone.

So you repeat with seven.

Eight, nine and ten are gone, so you go to eleven and repeate.

And you keep going.

I know you all know that. But I hope to explain that to someone some day.

Rohan Zener• Oct 15, 2016 #In decimal, i believe, that would have near about eight million digits, maybe more.

Xantiago P• Oct 15, 2016 #much numbers, such wow

Hedning1390• Oct 15, 2016 #You didn't remember the number in base two (or you forgot). I saw you sneak a peek at what you wrote above at 3:39.

TheAetherBros• Oct 15, 2016 #anynumer by the power of 2 – 1 is always a prime

TheAetherBros• Oct 15, 2016 #WTF IS THE DIFFERENCE BETWEEN GOOGLE AND GOOGOL

George Hansen• Oct 15, 2016 #You are a delightful and thoughtful way for me to start my week! Thank you for your enthusiasm and the information density of your videos! Your work is highly valued! Keep it up! We need you!! Sincerely, an old guy who still feels that math and arithmetic are poetry.