Ok, looking back at the original post I’m nearly regretting it as I need to find the time to actually write these – but at least this one was quick and simple. In fact it may be too simple to use as a tech test, but anyways here it is.

A prime number is any number greater than 1 which is only divisible by itself and 1. The method I use here simply checks all the values up to the number under investigation to see if division by them would leave a remainder of zero. There are probably faster ways to do this using smarter maths, and if you know of one, please post a comment!

` private static bool IsPrime(int num)`

{

if (num <= 1)

{

return false;

}

for (int i = 2; i < num; i++)

{

if (num % i == 0)

{

return false;

}

}

return true;

}

You can go twice as fast, if you have a separate test for divisibility by 2, and iterate over the odd numbers as test divisors.

You can finish your testing by an exponent of .5; namely, once your test divisor exceeds the square root of your candidate prime, you can terminate your loop.

If you are looking for clever maths, then see the "Agrawal–Kayal–Saxena primality test" as a general, polynomial, deterministic, and unconditional algorithm. If a candidate can explain it to you, then they are a "prime" candidate. 😉

Hello Dermot,

I'd like to propose another implementation:

private static bool IsPrime(uint n)

{

if (n == 0 || n == 1)

{

return false;

}

if (n == 2 || n == 3)

{

return true;

}

double squareRoot = Math.Sqrt(n);

for (uint i = 2; i <= squareRoot; i++)

{

if (n % i > 0)

{

return false;

}

}

return true;

}

The first part just check common cases. The second part, if the prime really exists then it should be num=p*q, then we should really check until the root square. Also I used uint (just positive integers).

Cheers,

Javier Andres Caceres Alvis

jacace.wordpress.com