What is a prime number?

A natural number greater than 1 with exactly two divisors: one and itself. The first primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.

Why is 1 not a prime number?

Because it has a single divisor — itself. If it counted as prime, prime factorisation would stop being unique: 6 = 2 · 3, but also 1 · 2 · 3, 1 · 1 · 2 · 3 and so on. Excluding one is what saves the fundamental theorem of arithmetic.

Yes — and it is the only even prime. Every other even number is divisible by 2, so it has at least three divisors and is composite.

How do you check whether a number is prime?

Look for divisors up to the square root of the number. If n is composite it has a divisor no greater than √n. For 91 we test 2, 3, 5, 7 (since 10² = 100 > 91) and find 7: 91 = 7 · 13, so 91 is not prime.

What is prime factorisation?

Writing a number as a product of primes only, e.g. 84 = 2² · 3 · 7. The fundamental theorem of arithmetic says such a factorisation exists for every number greater than 1 and is unique up to the order of the factors.

How many prime numbers are there?

Infinitely many — Euclid proved it. If the list of all primes were finite, the number equal to their product plus one would not be divisible by any of them, contradicting the assumption.

Intermediate level

Prime numbers

A prime number has exactly two divisors: one and itself. Learn the definition, the sieve of Eratosthenes, prime factorisation, and the fundamental theorem of arithmetic.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

GCD and LCMThe greatest common divisor and the lowest common multiple of two numbers. Learn the divisibility rules, the listing method, the Euclidean algorithm, and the formula tying GCD to LCM.

All formulas

Definition of a prime
$p \text{ is prime} \iff p > 1 \text{ and } p \text{ has exactly 2 divisors}$
its only divisors are 1 and itself
Fundamental theorem of arithmetic
$n = p_1^{a_1} \cdot p_2^{a_2} \cdot \ldots \cdot p_k^{a_k}$
the prime factorisation is unique
Primality test
$n \text{ composite} \implies \exists\, p \le \sqrt{n} : p \mid n$
divisors up to the square root are enough
GCD and LCM from factorisations
$\gcd = \prod p_i^{\min(a_i, b_i)}, \quad \operatorname{lcm} = \prod p_i^{\max(a_i, b_i)}$
lowest exponents for the GCD, highest for the LCM

A prime number is a natural number greater than $1$ with exactly two divisors: one and itself. A number greater than $1$ with more divisors is composite.

p \text{ is prime} \iff p > 1 \text{ and } p \text{ has exactly 2 divisors}

The list starts like this:

2, \; 3, \; 5, \; 7, \; 11, \; 13, \; 17, \; 19, \; 23, \; 29, \; 31, \; 37 \ldots

Two edge cases are worth remembering separately. One is not prime — it has a single divisor. Two is prime, and it is the only even prime: every other even number is divisible by $2$ and therefore has at least three divisors.

There are infinitely many primes — Euclid noticed that if the list were finite, the number equal to the product of all its entries plus $1$ would be divisible by none of them.

The sieve of Eratosthenes

The oldest way to list primes: write the numbers from $2$ to $n$ , then cross out the multiples of each number that survives. Only the primes remain.

For $n = 30$ : keep $2$ and cross out $4, 6, 8, \ldots$ ; keep $3$ and cross out $9, 15, 21, 27$ ; keep $5$ and cross out $25$ . Untouched are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29

The primality test — up to the square root

Testing a single number needs no sieve. It is enough to look for divisors up to the square root:

n \text{ composite} \implies \exists\, p \le \sqrt{n} : p \mid n

Why? Because if $n = a \cdot b$ and both factors were larger than $\sqrt{n}$ , their product would exceed $n$ . The smaller factor therefore always sits below the square root.

Is 91 a prime number?

Prime factorisation

Every natural number greater than $1$ can be written as a product of primes — and in only one way (up to the order of the factors). That is the fundamental theorem of arithmetic:

n = p_1^{a_1} \cdot p_2^{a_2} \cdot \ldots \cdot p_k^{a_k}

Primes are the "atoms" of arithmetic: every number is built from them, and in exactly one way.

The factorisation comes from dividing by successive primes, smallest first:

Factorise 84 into primes

GCD and LCM from factorisations

Factorisation gives a third way — next to listing divisors and the Euclidean algorithm — to compute the GCD and the LCM. Take the exponents of the shared primes: the lowest ones for the GCD, the highest ones for the LCM.

\gcd = \prod p_i^{\min(a_i, b_i)}, \qquad \operatorname{lcm} = \prod p_i^{\max(a_i, b_i)}

Compute the GCD and the LCM of 84 and 90 from their factorisations

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

smallest prime factor of 10

Common mistakes

Counting $1$ as prime — it has one divisor, and the definition asks for exactly two.
Ruling $2$ out for being even — being even is no obstacle; $2$ is prime, just the only even one.
Testing divisors all the way up to $n$ — up to $\sqrt{n}$ is enough, which cuts the work many times over.
Leaving a composite number in the factorisation — $84 = 4 \cdot 21$ is not a prime factorisation yet; carry on to $2^2 \cdot 3 \cdot 7$ .

Formula card

Topic: Prime numbers

Definition of a prime
$p \text{ is prime} \iff p > 1 \text{ and } p \text{ has exactly 2 divisors}$
its only divisors are 1 and itself
Fundamental theorem of arithmetic
$n = p_1^{a_1} \cdot p_2^{a_2} \cdot \ldots \cdot p_k^{a_k}$
the prime factorisation is unique
Primality test
$n \text{ composite} \implies \exists\, p \le \sqrt{n} : p \mid n$
divisors up to the square root are enough
GCD and LCM from factorisations
$\gcd = \prod p_i^{\min(a_i, b_i)}, \quad \operatorname{lcm} = \prod p_i^{\max(a_i, b_i)}$
lowest exponents for the GCD, highest for the LCM

Frequently asked questions

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