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#1 2014-06-28 05:27:14

Primenumbers
Member
Registered: 2013-01-22
Posts: 37

I need help with primes!

I need help with these prime numbers...

I can't seem to find them, what do you think of these ideas;

1. You multiply a prime by a certain number. That will somehow keep all remainders of p, a prime number, prime. Like 1.25 x a No. <25 etc.

2. You get groups of multiples of primes that do not align with each other. i.e. say you have 3, 7, 11 primes multiply each of them by separate multiples of 10(to the power of x) and they won't divide any of them by that No.

Please help I've been working on primes.

By basis for my theory is that a prime no. is not divisible by primes < sqrtp only. And primes if you work with the factors of primes you can generate them with various methods!

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#2 2014-06-28 06:10:45

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 88,475

Re: I need help with primes!

Hi Primenumbers;

For 1). Remainders of p? A remainder is what is left over when a number is divided by a another number. What number are you dividing by?


In mathematics, you don't understand things. You just get used to them.

I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.

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#3 2014-06-28 07:07:03

Primenumbers
Member
Registered: 2013-01-22
Posts: 37

Re: I need help with primes!

So 7 for example is not factorable by 2.
7/2=3 + remainder 1
because 7 is <9 I need only to see if it is divisible by 2 for it to not be prime.
<25, 3 <49,5 etc.

Last edited by Primenumbers (2014-06-28 07:08:06)

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#4 2014-06-28 13:05:53

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 88,475

Re: I need help with primes!

Hi;

Yes, you only have to check up to √n to test for primality of n. Did you want something else for 1)?


In mathematics, you don't understand things. You just get used to them.

I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.

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#5 2014-06-28 21:09:04

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,463

Re: I need help with primes!

hi Primenumbers,

By basis for my theory is that a prime no. is not divisible by primes < sqrtp only.

Let's say that xy = N.  If x = y then we call x the square root of N.

If x < √N, then y must be > √N in order that x times y still comes to N.  That is true even if x is not a +ve integer.

So if you're searching for factors of N, you only need to check from 2 up to √N.  If you fail to find one by then, you won't find one between √N and N either.

And you can shorten the search list further by not bothering to check non primes between 2 and √N.

reason:  if a non prime z divides N, then all the prime divisors of z will also divide N.  Hence it is sufficient just to check the primes.

So, if you have this list of primes between 2 and √N, and none of them divide N, then N had no divisors other than 1 and N, and so is a prime.

Your original post appears to ask more than that.  You'll have to give more detail if you still haven't got the answer you wanted.

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#6 2014-06-28 21:43:46

Primenumbers
Member
Registered: 2013-01-22
Posts: 37

Re: I need help with primes!

So if I minus multiples of primes that are

  from
until reaching other prime numbers, it proves
is prime........

Last edited by Primenumbers (2014-06-29 08:31:28)

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#7 2014-06-28 22:57:12

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,463

Re: I need help with primes!

eg.  Is 53 a prime ?

√53 = 7.*****

So Just need to try 2, 3 , 5 and 7.

As none are factors, => 53 is prime.

eg.  Is 48 prime ?

Note the factors come in pairs:

1,48
2,24
3,16
4,12
6,8

Once you have found a lower factor from {1,2,3,4,6} you automatically know the higher one of the pair {48,24,16,12,8} .

√48 = 6.****  ie. it lies between the closest pair, 6 < √48 < 8

The pairs are always one below and one above the square root ( unless the number is a perfect square of course which is why you need to consider the square root itself if it is a whole number )

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#8 2014-06-29 07:01:42

Primenumbers
Member
Registered: 2013-01-22
Posts: 37

Re: I need help with primes!

Let (n) = any number...
(n) - 2 - 2 -2 -2 -2 -2 -2 -2 ..............= any prime > 2
(n) - 3 -3 -3 -3 -3 -3 -3 -3 ...............= any prime > 3
(n) - 5 -5 -5 -5 -5 -5 -5 -5 ...............= any prime > 5
if continue to prime


this will prove the primality of

Last edited by Primenumbers (2014-06-29 08:44:09)

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#9 2014-06-29 20:52:43

Primenumbers
Member
Registered: 2013-01-22
Posts: 37

Re: I need help with primes!

Example 1:

= <11
109 - 2 = 107 (prime)
109 - 6 = 103 (prime)
109 - 20 = 89 (prime)
109 - 42 = 67 (prime)
therefore 109 is prime
Example 2:
= <11
119 - 6 = 113 (prime)
119 - 10 = 109 (prime)
119 - 112 = 7 (not prime>7)
therefore 119 is not prime
Example 3:
= <13
163 - 6 = 157 (prime)
163 - 50 = 113 (prime)
163 - 14 = 149 (prime)
163 - 66 = 97 (prime)
therefore 163 is prime

Last edited by Primenumbers (2014-06-29 20:55:37)

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