# Prime & Composite Numbers

Definition: A prime number is a whole number with exactly two integral divisors, $1$ and itself.

The number $1$ is not a prime, since it has only one divisor.

So the smallest prime numbers are:

$2,3,5,7,\cdots$

The number $4$ is not prime, since it has three divisors ( $1$ , $2$ , and $4$ ), and $6$ is not prime, since it has four divisors ( $1$ , $2$ , $3$ , and $6$ ).

Definition: A composite number is a whole number with more than two integral divisors.

So all whole numbers (except $0$ and $1$ ) are either prime or composite.

Example:

$43$ is prime, since its only divisors are $1$ and $43$ .

$44$ is composite, since it has $1,2,4,11,22$ and $44$ as divisors.

## How can you tell if a number is prime?

First of all, here are some ways to tell if a number is NOT prime:

Any number greater than $2$ which is a multiple of $2$ is not a prime, since it has at least three divisors: $1$ , $2$ , and itself. (This means $2$ is the only even prime.)

Any number greater than $3$ which is a multiple of $3$ is not a prime, since it has $1$ , $3$ and itself as divisors. (For example, $303$ is not prime, since $303÷3=101$ .)

Any number which is a multiple of $4$ is also a multiple of $2$ , so we can rule these out.

Any number greater than $5$ which is a multiple of $5$ is not a prime. (So the only prime number ending with a $0$ or $5$ is $5$ itself.)

Any number which is a multiple of $6$ is also a multiple of $2$ and $3$ , so we can rule these out too.

You can continue like this... basically, you just have to test for divisibility by primes!

Example 1:

Is $119$ prime?

First test for divisibility by $2$ . $119$ is odd, so it's not divisible by $2$ .

Next, test for divisibility by $3$ . Add the digits: $1+1+9=11$ . Since $11$ is not a multiple of $3$ , neither is $119$ . (Remember, this trick only works to test divisibility by $3$ and $9$ .)

Since $119$ doesn't end in a $0$ or a $5$ , it's not divisible by $5$ .

Next, test for divisibility by $7$ . You'll find that $119÷7=17$ .

So the answer is NO... $119$ is not prime.

Example 2:

Is $127$ prime?

First test for divisibility by $2$ . $127$ is odd, so it's not divisible by $2$ .

Next, test for divisibility by $3$ . Add the digits: $1+2+7=10$ . Since $10$ is not a multiple of $3$ , neither is $127$ .

Since $127$ doesn't end in a $0$ or a $5$ , it's not divisible by $5$ .

Next, test for divisibility by $7$ . You'll find that $7$ doesn't go in evenly.

The next prime is $11$ . But $11$ doesn't go in evenly, either.

You can stop now... it must be prime! You don't need to keep checking for divisibility by the next primes ( $13,17,19,23,$ etc.). The reason is that if $13$ went in evenly, then we would have $127=13×n$ for some number $n$ . But then $n$ would have to be less than $13$ ... and we already know that $127$ is not divisible by any number smaller than $13$ .

So the answer is YES... $127$ is prime.

For more advanced topics and a list of the first $400$ primes, go to the Prime Page or the page on prime factorization .