1.1 Prime Numbers and Composite Numbers

Every whole number greater than $1$ is either prime or composite, and you tell them apart by COUNTING factors. A prime has exactly two, $1$ and itself. A composite has more than two. The count is what matters: $1$ has only one factor, so it is neither.

The first primes are $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, \ldots$

Classifying numbers as prime or composite

State whether each of $1$ , $2$ , $9$ , $23$ and $51$ is prime, composite, or neither. Give a reason each time.

$1$ : its only factor is $1$ , so it has ONE factor. It is neither prime nor composite.
$2$ : factors are $1$ and $2$ , exactly two, so it is prime. It is the only even prime.
$9$ : factors are $1, 3, 9$ , so it is composite ( $9 = 3 \times 3$ ).
$23$ : no prime up to $\sqrt{23} \approx 4.8$ divides it, so testing $2$ and $3$ is enough. It is prime.
$51$ : $5 + 1 = 6$ is divisible by $3$ , so $3 \mid 51$ and $51 = 3 \times 17$ . It is composite.

Try it

A practice question on this appears here. Practise primes hcf lcm.

A number with exactly 3 factors

A whole number has exactly 3 factors. Show that it must be the square of a prime number, and find the smallest such number.

If $n = p^2$ where $p$ is prime, then the factors of $n$ are $1, p, p^2$ , which is exactly 3.
Every other shape gives a different count: $n = 1$ has 1 factor; a prime $n = p$ has 2; a prime power $n = p^k$ with $k \geq 3$ has at least 4 ( $1, p, p^2, p^3$ ); and any $n$ with two different prime factors has at least 4 ( $1, p, q, pq$ ).
So $n = p^2$ is the only possibility.
The smallest prime is $2$ , so the smallest such number is $2^2 = 4$ (factors: $1, 2, 4$ ).

Practise this

Questions on Primes, HCF and LCM, marked as you go, with the working shown.

Practise Primes, HCF and LCM