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Primenumbers

L

Lillen

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I believe that the primes up to 19 (2,3,5,7,11,13,17,19,) can be used to calculate primes above those number, even 10 digit primes, in your head. If the ten digit number cannot be devided with those eight numbers it is a prime.

I have had this discussion before, were those who argued against me said that i was wrong because if you multply primes above 19 with eachother you will have a number that isn't fit in to those eight multiplication tables.

The error in that argument was that they are looking at numbers that aint a prime. For instance 23*29 = 667 and 667 is not counted in those eight multiplication tables. But it is not a prime.

My theory or axiom (whatever I am not a scientist) is that if you have a ten digit prime, all you need to do is check with those eight primes to verify if your ten digit prime is a prime.

I am uncertain of this, and it needs to be discussed.
Arguments for it is that those eight primes multiplication tables do not cover no other primes, and the product of two primes above 19.

If this can be varified or if it is set aside as a foul idea is yet to be discovered. If it is verified to bruteforce primes will be easier since you take x/y were x is the unknowned prime, and y is one of the eight primes established.

As we all know primes can only be devided with one and itself. But instead of using a bruteforce that calculate each and every digit there is, my hypthesis provides us with only eight numbers to verify the unknowned prime.

I am willing to take critics towards this method, and I am not certain that my method works...
 

Delphiki

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I believe that the primes up to 19 (2,3,5,7,11,13,17,19,) can be used to calculate primes above those number, even 10 digit primes, in your head. If the ten digit number cannot be devided with those eight numbers it is a prime.

A prime number is any number that has no divisors other than 1 and itself. Any number that can't be divided by anything other than 1 or itself is considered prime.
 
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Freodin

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Your idea is incorrect. I´ll if I can make the proof understandable.

Every number can be written as the multiplication of prime numbers. In German, this is called "Primfaktorenzerlegung" (division into prime factors).

So if a number is a product of prime factors other than your mentioned ten, it cannot be a prime number, but is not divisible by your set numbers.

q.e.d.

If the ten digit number cannot be devided with those eight numbers it is a prime.
...
For instance 23*29 = 667 and 667 is not counted in those eight multiplication tables. But it is not a prime.
You have already found the problem with your hypothesis yourself. It seems you just didn´t understand it.

Your first statement is that every such number that is not divisible by your set numbers is a prime. But your second statement explicitly gives such a number that is not divisible by your set, and is not a prime. You have disproven yourself.
 
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Lillen

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Ok so it can give outcomes for either the product of two primes, or a prime.

Lets look at another problem i need you lifescientists to explain and clarify for me:

How do i write the following sentence properly:

6x^2+1^2+y=0

Can you write it that way???

Man it was along time ago i read this in school, I have been depoisoned atleast 8 times without doing drugs. This made me cleanse my mind from all i've learnt.

I read about these kinds of equations 8 years ago and got a B+ (VG in swedish), the second highest grade. I used to be really good at highschool math. It was when we came to college math, trignometery, i stopped basically because I felt that i had to go back to basics again, due to the brainwashing on institutions...
 
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Freodin

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Ok so it can give outcomes for either the product of two primes, or a prime.

Lets look at another problem i need you lifescientists to explain and clarify for me:

How do i write the following sentence properly:

6x^2+1^2+y=0

Can you write it that way???
That depends on what you want to say.

a x^2 + b x + c = 0

This is the common syntax for a equation of the second degree. Depending on the values for a, b and c is has none, one or two possible solutions for x, which can be computed by a set formula.

An equation using x and y usually has both these variables unknown... such an equation has no explicit solutions on its own. You´d need a second equation to solve such a system.
 
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Upisoft

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I have had this discussion before, were those who argued against me said that i was wrong because if you multply primes above 19 with eachother you will have a number that isn't fit in to those eight multiplication tables.

The error in that argument was that they are looking at numbers that aint a prime. For instance 23*29 = 667 and 667 is not counted in those eight multiplication tables. But it is not a prime.
It is not error, it is called counterexample. It shows the error in your logic.
 
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Freodin

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6x^2 + 1^2 + 187 = 0

can you come to the roots 11 and 17 from there?

Huh? I´m sorry, I cannot follow you. Language problem, I guess... I simply don´t understand what you are asking.

Let´s see: what you wrote can be simplified. 1^2 is still 1. No need to seperate this 1 from the 187 also, so we get to...

6x^2 + 188 = 0

Further transformation gets to...

x^2 = -94/3

As it is impossibel for a square of a real number to be negative, there is no solution for this equation in this group.

That said... perhaps you can explain what you are trying to say?
 
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JustMeSee

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6x^2 + 1^2 + 187 = 0

can you come to the roots 11 and 17 from there?
Do you mean:

(6x^2) + (x) + 187 = 0 ?

That would involve factoring or the quadratic equation to solve.

(-b + sqrt( b^2 - 4ac)) / 2a
or
(-b - sqrt( b^2 - 4ac)) / 2a

a = 6
b = 1
c = 187

This would then be:
(-1 + sqrt(1^2 - 4*6*187))/ (2*6)
or
(-1 - sqrt(1^2 - 4*6*187))/ (2*6)

You CAN NOT solve for x in this equation because sqrt(1^2 - 4*6*187) = sqrt(-4487). This would be an imaginary number.

IF the equation was:

(6x^2) + (x) - 187 = 0


(-b + sqrt( b^2 - 4ac)) / 2a
or
(-b - sqrt( b^2 - 4ac)) / 2a

a = 6
b = 1
c = -187


It could then be solved for:

(-1 + sqrt(4489))/12 = -5.5

or


(-1 - sqrt(4489))/12 = 5.6667

x= -5.5 or 5.6667

I am not sure if this is the equation/formula you were looking for.
 
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JustMeSee

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First 100 prime numbers:
100.gif


http://www.math.uic.edu/~lewis/las100/primes.html
 
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Wiccan_Child

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Whee, maths!

I believe that the primes up to 19 (2,3,5,7,11,13,17,19,) can be used to calculate primes above those number, even 10 digit primes, in your head. If the ten digit number cannot be devided with those eight numbers it is a prime.
What about 667? It's not divisible by either 2, 3, 5, 7, 11, 13, 17, or 19, so your system would conclude that it's a prime... but 667 is divisible by both 23 and 29.

The error in that argument was that they are looking at numbers that aint a prime. For instance 23*29 = 667 and 667 is not counted in those eight multiplication tables. But it is not a prime.
Well, yea, that's the point: your system says it's a prime, but it's not. Thus, your system is wrong.

How do i write the following sentence properly:

6x^2+1^2+y=0

Can you write it that way???
Sure. But conventionally it would be written 6x[sup]2[/sup] + y + 1 = 0.

6x^2 + 1^2 + 187 = 0

can you come to the roots 11 and 17 from there?
Do you mean, are te roots of that equation 11 and 17? Well, as Freodin said, that equation doesn't have real roots.
 
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L

Lillen

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First of, It was long time ago i read about these kinds of equation.

I said it WC, there are exceptions for my formula, and the exceptions are the product of two primes. You cant devide that either with those numbers.

But Vet came up with another system, multply 6 with any number and minus or add (some of it) with one and you will have a primenumber.

I checked and in this system you can end up with a product of two primes as well.. there exceptions so to say..

What I´m trying to do is set up a formula so that you can figure out two diffrent primes only knowing their product.

I've said this on other boards (utopiaforums) and i believe that it is possible doing so only using highschool math...

You cant square negtive numbers

Thanks for the reminder
 
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Freodin

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First of, It was long time ago i read about these kinds of equation.

I said it WC, there are exceptions for my formula, and the exceptions are the product of two primes. You cant devide that either with those numbers.
That again is not quite correct.
Every number is the product of prime numbers. For example: 215016 = 2*2*2*3*17*17*31

That means that your system does not work for any number that is not a product of only the first ten prime numbers.

Example: 35557. It is not divisible by the first ten primes. It is also not the products of two primes... it is the product of three primes.

So to find out if a number is a prime number, you have to check. You have to check if they are dividable by a prime number. Starting with the first ten... and then up to the prime number closest to the squareroot of the number to be tested.

And this is exactly the way it is done now. Your method does not provide any advantage.

But Vet came up with another system, multply 6 with any number and minus or add (some of it) with one and you will have a primenumber.

I checked and in this system you can end up with a product of two primes as well.. there exceptions so to say..
And to find if your test number is such an exception or not, you can do nothing but follow the classical method of testing one divisor after the other. So AV´s method is as redundant as yours. It does not provide any additional informations.

It is a wonderful example of a system that is always right... except when it is wrong.

What I´m trying to do is set up a formula so that you can figure out two diffrent primes only knowing their product.
I´d say that if this thread has shown something, it is that your mathematical knowledge is - for whatever reasons - seriously lacking.

So if you are going to fathom one of the most difficult problems in mathematics - factorization algorithms - while not even understanding primefactors... you could colour me really impressed.

Integer factorization - Wikipedia, the free encyclopedia

I've said this on other boards (utopiaforums) and i believe that it is possible doing so only using highschool math...
Sure....


Thanks for the reminder
I don´t know who you quoted here... I can´t find anyone in this thread having said that. So I guess when you quoted "You can´t square negative numbers" you were stating your understanding of what we tried to tell you.

Well, you understood it wrong: There is no problem with squaring a negative number... but you will never get a negative result from squaring any real number.
 
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Wiccan_Child

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First of, It was long time ago i read about these kinds of equation.

I said it WC, there are exceptions for my formula, and the exceptions are the product of two primes. You cant devide that either with those numbers.
But every number is a product of primes. Every number that isn't a prime number can be written as a product of primes: 10=2x5, 169=13x13, 360=2x2x2x3x3x5, etc. Any number that is a product of primes not in your list, invalidates your system. They're not just exceptions, they're glaring holes.

But Vet came up with another system, multply 6 with any number and minus or add (some of it) with one and you will have a primenumber.

I checked and in this system you can end up with a product of two primes as well.. there exceptions so to say..

What I´m trying to do is set up a formula so that you can figure out two diffrent primes only knowing their product.
By definition, primes don't have a product.

I've said this on other boards (utopiaforums) and i believe that it is possible doing so only using highschool math...
Primes are one of the most mysterious categories of number. An exhaustive iterative sequence is not trivial. The Reimann hypothesis, though unproven, is the closest you'll get.
 
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Upisoft

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What I´m trying to do is set up a formula so that you can figure out two diffrent primes only knowing their product.
If you do that CIA will be very happy. You know, they want to be able to easily break RSA encryption... which is based on the fact we have no easy way to do what you want to do at the moment. However when we have quantum computing RSA will be history.
 
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