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shahzadjalbani

Is there any equation to calculate the prime numbers?

  • one year ago
  • one year ago

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  1. nbouscal
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    Nope

    • one year ago
  2. lgbasallote
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    do you mean a formula to verify if it is prime?

    • one year ago
  3. shahzadjalbani
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    Yes.....

    • one year ago
  4. nbouscal
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    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

    • one year ago
  5. nbouscal
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    Sieve of Eratosthenes is a good place to start if you want to research this area

    • one year ago
  6. lgbasallote
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    i can come up with a formula impromptu....o.O

    • one year ago
  7. lgbasallote
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    but yeah...there isnt any im familiar of

    • one year ago
  8. ParthKohli
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    Easiest way: check if it's divisible by any natural number or not :P

    • one year ago
  9. nbouscal
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    Trust me the number theorists would love if there was an easy formula to verify primality :P

    • one year ago
  10. nbouscal
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    So far, there isn't. If there was the Reimann Hypothesis wouldn't be as big of a deal as it is.

    • one year ago
  11. ParthKohli
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    I see something interesting. http://en.wikipedia.org/wiki/Lucas%E2%80%93Lehmer_primality_test

    • one year ago
  12. lgbasallote
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    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

    • one year ago
  13. nbouscal
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    Yeah parth that is for Mersenne primes only, though

    • one year ago
  14. ParthKohli
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    ^ spam. It didnt help

    • one year ago
  15. lgbasallote
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    are you good now @shahzadjalbani ?

    • one year ago
  16. shahzadjalbani
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    @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 .

    • one year ago
  17. nbouscal
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    There is not a simple way with the primes. The primes are not simple. That's why people are still studying them :)

    • one year ago
  18. lgbasallote
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    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

    • one year ago
  19. lgbasallote
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    i have no way of knowing because the number could be divisible by the succeeding prime numbers or it could be just prime

    • one year ago
  20. nbouscal
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    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.

    • one year ago
  21. lgbasallote
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    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

    • one year ago
  22. shahzadjalbani
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    @nbouscal explain your statement.

    • one year ago
  23. nbouscal
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    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.

    • one year ago
  24. lgbasallote
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    or just divide by the primes

    • one year ago
  25. nbouscal
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    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.

    • one year ago
  26. lgbasallote
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    20/2 wields a quotient with no remainder so it is automatically not prime

    • one year ago
  27. nbouscal
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    Obviously 20 isn't prime, lgba, that was not the point of the example.

    • one year ago
  28. lgbasallote
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    i was giving an eample too...of my method

    • one year ago
  29. nbouscal
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    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.

    • one year ago
  30. lgbasallote
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    makes sense...

    • one year ago
  31. nbouscal
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    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.

    • one year ago
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