![]() It will only test the groups of (5, 7), (11, 13) and so on. Therefore, it skips testing numbers that are divisible by 2 or 3. ![]() It uses the fact that all primes (except 2 and 3) have to be at 6k☑.So if I am given a number, say n, than I can check in the range (n, 2n. A weaker but more elegant formulation is: for every n > 1 there is always at least one prime p such that n < p < 2n. However, most people want prime calculations to be fast, which is more important than using the Stream API what would bring huge overhead. Bertrands postulate (actually a theorem) states that if n > 3 is an integer, then there always exists at least one prime number p with n < p < 2n 2. The original post likes to use Java 8's Stream API.It is sufficient to do the check up to Math.sqrt(candidate).The solution is faster than other boolean isPrime(int candidate) solutions due to: See Wikipedia: Ī prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. The solution in the original post, the accepted answer, and many more answers are incorrect because they identify negative numbers, zero, and one as prime numbers, which they are not. I am coming late to the game, but I am sharing my solution because there is a surprising number of incorrect implementations shared and the following code has better performance than the other boolean isPrime(int candidate) solutions.
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