**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.

**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 $303\xf73=101$.)

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 $119\xf77=17$.

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 $127=13\times n$ 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.