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shahzadjalbani
Is there any equation to calculate the prime numbers?
do you mean a formula to verify if it is prime?
There are a lot of methods for finding prime numbers, but it's not that easy, there's no function to check if any number is prime without having a list of prime numbers to use
Sieve of Eratosthenes is a good place to start if you want to research this area
i can come up with a formula impromptu....o.O
but yeah...there isnt any im familiar of
Easiest way: check if it's divisible by any natural number or not :P
Trust me the number theorists would love if there was an easy formula to verify primality :P
So far, there isn't. If there was the Reimann Hypothesis wouldn't be as big of a deal as it is.
I see something interesting. http://en.wikipedia.org/wiki/Lucas%E2%80%93Lehmer_primality_test
here is a list of *some* prime numbers if itll help: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997
Yeah parth that is for Mersenne primes only, though
^ spam. It didnt help
are you good now @shahzadjalbani ?
@lgbasallote You sent a list of numbers now I have to compare each number with this list it will more difficult and lengthy job . Is there any simple way to do it .
There is not a simple way with the primes. The primes are not simple. That's why people are still studying them :)
my way is bloodily and manually divide a number by the simplest prime numbers (2, 3, 5, 7, 11) if i get a quotient with no remainder it is not prime....if there is a remainder...i have no way f knowing so i give up and say true lol
i have no way of knowing because the number could be divisible by the succeeding prime numbers or it could be just prime
One important method when determining if a number is prime is to bound the potential divisors by the square root of the number you're investigating. Once you hit that bound, you know it's prime.
but normally...if the number is not divisible by the basic prime numbers then it is prime...teachers are not that sadistic to have you try each one
@nbouscal explain your statement.
Well, any integer divisors come in pairs, right? So if you're looking at the number 20, for example, you have pairs 1,20; 2, 10; 4, 5... then you have 5,4; 10,2; 20,1... but you already had those, just backwards. So if you're checking primes, and you check 1, 2, 4... you don't need to check any more numbers past that because you already would have found it in those first pairs. Once you hit the square root of the number, you know you don't have to check anymore.
or just divide by the primes
So for example you want to check if 51 is prime, and you know that sqrt(51) is just a bit over 7, you only have to check if 51 is divisible by 2, 3, 5, 7. You don't need to go any further than that.
20/2 wields a quotient with no remainder so it is automatically not prime
Obviously 20 isn't prime, lgba, that was not the point of the example.
i was giving an eample too...of my method
The point was to show why you don't need to worry about primes greater than the square root, because you would already have run into them before you got to the square root.
The square root bound is very useful when you are using computer methods to check primality on large numbers, it saves a lot of time.