## Standing ovation puzzle

For fans of all other kinds of logic puzzles

### Standing ovation puzzle

In 1903 a man gave a wordless lecture to a prestigious American society where he wrote a 12 digit number on a blackboard and multiplied it by a 9 digit number. When he had finished, he received a standing ovation. Who was he, what was the society, and why was he applauded?
rjamil

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Location: Karachi, Pakistan

### Re: Standing ovation puzzle

F.N cole
American mathematical society

2^67 - 1
Is what he calculated out by hand as evedince it was not. prime number.

Fixed the typo
Lol
Last edited by StrmCkr on Wed Aug 15, 2018 10:19 pm, edited 1 time in total.
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StrmCkr

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### Re: Standing ovation puzzle

StrmCkr wrote: .... as evedince it was a prime number.

You must mean "as evidence it was NOT a prime number".

Most numbers of the form 2^P - 1, where P is prime, are themselves prime. 2^67 - 1 must be one of the rare exceptions.

Bill Smythe
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### Re: Standing ovation puzzle

Derp, that's what I get for autocorrect. Changing and deleting words. On my phone...
Some do, some teach, the rest look it up.

StrmCkr

Posts: 1205
Joined: 05 September 2006

### Re: Standing ovation puzzle

I think I remember reading, in a recreational math book I checked out from the school library when I was about 12, that it had once been conjectured that whenever P is prime, 2^P - 1 is also prime. The book also said that the conjecture was eventually disproved via a counterexample. I guess that's the one you showed.

Is 2^67 - 1 the first (i.e. the lowest) counterexample? Does anyone know what the next one is?

Bill Smythe
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### Re: Standing ovation puzzle

the lowest counterexample is 2^11 - 1 = 2047 = 23 * 89

mersenne primes of the form 2^p-1 become pretty rare as you go up, but still account for the biggest prime numbers found today

back in the 1800's a lot of recreational mathematicians devoted their time to factorising (or finding prime) numbers of the form 2^n +- 1, it was the fashionable thing of the day for people to do, no idea why I guess people just liked big prime numbers

my favourite related story to tell when this topic comes up is about that guy (Landry?) who devoted his entire life to doing this and got stuck on 2^58+1 which factorises as 5 x big number x big number , and when he got there he pointed out how frustratingly hard it was to do and how many years of hard labour it took for him to do it

then shortly after this was published, aurifeuille points out that
4n^4 + 1 = (2n^2+2n+1) * (2n^2 - 2n + 1)

so this guy put in all these years of effort for no real reason. i'm on mobile but if i wasn't I'd link to sources, but i'm sure you can look this up

I feel like this needs to be made into a meme
999_Springs

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Location: In the toilet, flushing down springs, one by one.

### Re: Standing ovation puzzle

.

The biggest prime known is, I believe, 2 ^ 77,232,917 - 1

You can get info on the great Mersenne prime hunt at https://www.mersenne.org/

There is a reason why the largest primes found to date are Mersenne primes. Mersenne numbers (ie any number of the form 2^N - 1) are easier to test for primality than other numbers, owing to some tricks that can only be applied to Mersenne numbers.

I'm actually playing with an idea for finding large primes other than the Mersenne primes, but I'm probably just playing with myself for even thinking it is possible

Mathimagics
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### Re: Standing ovation puzzle

Hi Mathimagics,

Mathimagics wrote:I'm actually playing with an idea for finding large primes other than the Mersenne primes, but I'm probably just playing with myself for even thinking it is possible

According to the Wikipedia, A prime number (or a prime) is a natural number greater than 1 that cannot be formed by multiplying two smaller natural numbers. A natural number greater than 1 that is not prime is called a composite number.

I don't know why below idea comes in to my mind:
A prime number (or a prime) is a natural number greater than 1 that cannot be divisible by smaller prime numbers.

Anybody either confirm or refute that I might be wrong?

R. Jamil

Volunteers have a chance to earn research discovery awards of \$3,000 or \$50,000 if their computer discovers a new Mersenne prime. GIMPS' next major goal is to win the \$150,000 award administered by the Electronic Frontier Foundation offered for finding a 100 million digit prime number.
rjamil

Posts: 601
Joined: 15 October 2014
Location: Karachi, Pakistan

### Re: Standing ovation puzzle

A prime number (or a prime) is a natural number greater than 1 that cannot be divisible by smaller prime numbers.

Anybody either confirm or refute that I might be wrong?

a prime number cannot be divided by any number but its self and 1.
other wise it is also a composite number....

some fun with numbers
a series of numbers raised to an equal power and added together

where N = 0=>> infinity

0:(1^n + 2^n) = 2
1:(1^n + 2^n) = 3
2:(1^n + 2^n ) = 5 =>>> (1^(1^n + 2^n ) + 2^(1^n + 2^n ) ) { 2 = (1^n + 2^n) replace all n's with the equation for "0" and use n as 0 and surprise the answer is still 5.
3:(1^n + 2^n) = 9 =>> (1^(1^n + 2^n) + 2^(1^n + 2^n)) { 3 = (1^n + 2^n) replace all n's with the equation for "1" and use n as 1 and surprise the answer is still 9.
4:(1^n + 2^n) = 17
5:(1^n + 2^n) = 33 =>> (1^(1^n + 2^n )+ 2^(1^n + 2^n )) {5 = (1^n + 2^n ) replace all n's with the equation for "2" and use N as 2 and surprise the answer is still 33},
{goofier yet 2 = (1^n + 2^n ) replace all n's with the equation for "2" and use N as 0 and surprise the answer is still 33.
=>> (1^(1^(1^n + 2^n) + 2^(1^n + 2^n) )+ 2^(1^(1^n + 2^n) + 2^(1^n + 2^n) ))
----------------------------------------------
0:(1^n + 3^n) = 2
1:(1^n + 3^n) = 4
2:(1^n + 3^n ) = 10
3:(1^n + 3^n ) = 28
4:(1^n + 3^n ) = 82
5:(1^n + 3^n ) = 244
-------------------------------------------------
0:(1^n + 4^n) = 2
1:(1^n + 4^n) = 5 -> prime
2:(1^n + 4^n ) = 17 -> prime
3:(1^n + 4^n ) = 65
4:(1^n + 4^n ) = 257
5:(1^n + 4^n ) = 1025
-----------------------------------------
0:(1^n + 5^n) = 2
1:(1^n + 5^n) = 6
2:(1^n + 5^n ) = 26
3:(1^n + 5^n ) =126
4:(1^n + 5^n ) = 626
5:(1^n + 5^n ) =3126
----------------------------
0:(2^n + 2^n) = 2
1:(2^n + 2^n) = 4
2:(2^n + 2^n ) = 8
3:(2^n + 2^n ) = 16
4:(2^n + 2^n ) = 32
5:(2^n + 2^n ) = 64
----------------------------
0:(2^n + 3^n) = 2 -> prime
1:(2^n + 3^n) = 5 -> prime
2:(2^n + 3^n ) = 13 -> prime
3:(2^n + 3^n ) = 35 -> prime
4:(2^n + 3^n ) = 97 -> prime
5:(2^n + 3^n ) = 275
----------------------------
0:(2^n + 4^n) = 2
1:(2^n + 4^n) = 6
2:(2^n + 4^n ) = 20
3:(2^n + 4^n ) = 72
4:(2^n + 4^n ) = 272
5:(2^n + 4^n ) =1056
----------------------------
0:(2^n + 5^n) = 2
1:(2^n + 5^n) = 7 -> prime
2:(2^n + 5^n ) = 29 -> prime
3:(2^n + 5^n ) = 133
4:(2^n + 5^n ) = 641 -> prime
5:(2^n + 5^n ) =3157
6:(2^n + 5^n ) =15689
------------------------------------

if u might have noticed there is a pattern here that can generate potentially all primes.
gl deciphering it.
Last edited by StrmCkr on Sun Sep 16, 2018 8:16 am, edited 1 time in total.
Some do, some teach, the rest look it up.

StrmCkr

Posts: 1205
Joined: 05 September 2006

### Re: Standing ovation puzzle

Hi,

StrmCkr wrote:a prime number cannot be divided by any number but its self and 1.
other wise it is also a composite number....

True. But it also means that a prime number has no smaller prime numbers factors.

I see no authentic statement about that there are no factorization of prime numbers except 1 and itself.

Since, it is clear that 2 is the only even number that is also a prime member, therefore, rest of the even numbers are composite numbers.

So, if we want to prove some very large odd natural number, whether it is also a prime number or a composite number, then we only need to divide that odd natural number by prime numbers between 3 and less then or equals to square root of that odd natural number.

- Apart from 2 and 5, all prime numbers have to end in 1, 3, 7 or 9 so that they can’t be divided by 2 or 5.

- But after devising a computer programme to search for the first 400 billion primes, the two mathematicians found prime numbers tend to avoid having the same last digit as their immediate predecessor – as if, in the words of Dr Lemke Oliver they “really hate to repeat themselves.”

R. Jamil
Last edited by rjamil on Fri Sep 21, 2018 7:41 pm, edited 1 time in total.
rjamil

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Location: Karachi, Pakistan

### Re: Standing ovation puzzle

rjamil wrote:So, if we want to prove some very large odd natural number, whether it is also a prime number or a composite number, then we only need to divide that odd natural number by prime numbers between 3 and less then or equal to square root of that odd natural number.

"Only"? Consider the number given above, 2 ^ 77,232,917 - 1. It has 23 million digits, so its square root has roughly 12 million digts. Even if one had all the primes up to 12 million digits, which we do not (and couldn't possibly store them anyway in all the computers on the planet), proving primality by this method (ie trial-division) would take billions of years.

That's even longer than the estimated P&P solving time for some of my Kakuro puzzles!

Mathimagics
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### Re: Standing ovation puzzle

where N: uses numbers < 5 and are primes, however they are not = to the difference of K-j or = J+K
where J is 1 -> all primes {can only be primes and primes^ of another prime power}
where K is 2 -> and all primes {can only use primes and primes^ of another prime power}

if J is even then k must be odd
if j is odd then k must be even

Q can be any prime number <5

((J^q)^n +( K^q)^n) => a new prime number.

from the above stuff we can convert this:
2:(1^n + 4^n ) = 17 -> prime
=>> ((1^2)^2 + (2^2)^2) =>> (1^2) + (4^2) = 17

anyway its way to late for random rambling... lol
Some do, some teach, the rest look it up.

StrmCkr

Posts: 1205
Joined: 05 September 2006

### Re: Standing ovation puzzle

Hi all,

Googling for prime number generator and found this site that provide up to 64bit program to generate 367783654 primes less than 8000000000 and took around 1 1/2 hours to print.

Maybe, someone will convert it into 128bit or more and then it will generate much more primes within reasonable time.

R. Jamil
rjamil

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