Determine Prime Number: Simple Algorithm w/ While Loop

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A simple algorithm to determine if a number is prime involves checking for divisibility by integers up to the square root of the number. The process includes testing only odd numbers, as even numbers greater than two cannot be prime. If a divisor is found that divides the number evenly, it is not prime; if no divisors are found, the number is prime. Using a remainder operator can simplify this check, allowing for efficient determination of primality. This method balances simplicity and efficiency in prime number identification.
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can smby give me a simple example algorithm to determine whether the number is prime or not such as using while loop, if...then...else or others which is simple

thanx
 
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but I'm looking for a simple example about prime number determination [algorithm] for programming language only using if...then... or while loop or repeat loops... that describe the steps of calculation
 
teng125 said:
but I'm looking for a simple example about prime number determination [algorithm] for programming language only using if...then... or while loop or repeat loops... that describe the steps of calculation
How many of those google hits did you actually look through?
 
A simple example I can think of is go through every integer between 1 and the prime number you want to test and check to see if it is a divisor.
 
buddyholly9999 said:
A simple example I can think of is go through every integer between 1 and the prime number you want to test and check to see if it is a divisor.

you only need to go up to a third of the number. This is because if it is even then you know it is not prime, unless it is two. odd numbers are not divisble by 2, so the next divisor would be 3.
 
a_ramage_1989 said:
you only need to go up to a third of the number. This is because if it is even then you know it is not prime, unless it is two. odd numbers are not divisble by 2, so the next divisor would be 3.

It's enough to go up to the square root of the number, much more efficient than having to check up to a third (still not at all fast though).
 
and you only need to test (up to the \sqrt{N} ) the odd numbers that are not known multiples of other prime numbers (like multiples of 3, 5, 7, 11, 13, etc). if your programming language has a remainder operator (like the "%" operator in C and C++), you can use that to test a candidate factor to see if it divides evenly into N. but if you don't, you have to divide, use the floor() or int() operator to get the integer part of the quotient, subtract the integer part from the quotient and, if that result is zero, it divides evenly.

if all candidate factors fail to divide evenly into N, then N is prime. if anyone candidate factor divides evenly into N, it is not prime.
 
okok...thanx
 
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