Prime factorisation and the Fundamental Theorem

The building blocks

A prime number has exactly two factors: $1$ and itself. The first primes are $2, 3, 5, 7, 11, 13, 17, 19, 23$. A whole number greater than $1$ that is not prime is composite. It can be split into smaller factors.

The number $1$ has only one factor, so it is neither prime nor composite.

Keep splitting a composite number, and you always end with primes. Primes are the building blocks of all whole numbers, as atoms are the building blocks of matter.

A factor tree

Write the number at the top. Split it into two factors. Split again any factor that is not prime. Stop when every branch ends in a prime.

So $840 = 2 \times 2 \times 2 \times 3 \times 5 \times 7 = 2^3 \times 3 \times 5 \times 7$. This is the prime factorisation of $840$. We write repeated primes as powers, smallest prime first.

Repeated division

The same work, set out in a column, is often quicker. Divide by $2$ as many times as you can, then by $3$, then by $5$, and so on.

• $1260 \div 2 = 630$
• $630 \div 2 = 315$
• $315 \div 3 = 105$
• $105 \div 3 = 35$
• $35 \div 5 = 7$, and $7$ is prime.

So $1260 = 2^2 \times 3^2 \times 5 \times 7$.

A short cut. You only need to try primes $p$ with $p \times p$ not more than what is left. If no such prime divides it, what is left is itself a prime. For $91$, try $2, 3, 5, 7$: $91 = 7 \times 13$. For $97$, none of $2, 3, 5, 7$ divides it, and $11 \times 11 = 121$ is already too big. So $97$ is prime.

The Fundamental Theorem of Arithmetic

Two students factorise $840$. One starts with $840 = 8 \times 105$, the other with $840 = 20 \times 42$. They follow different trees. Yet both end with the same primes: three $2$s, one $3$, one $5$ and one $7$.

This always happens. It is so important that it has a grand name.

The Fundamental Theorem of Arithmetic. Every composite number can be written as a product of primes, and this product is unique, apart from the order of the primes.

Now you can see why $1$ is not called a prime. If it were, we could write $6 = 2 \times 3 = 1 \times 2 \times 3 = 1 \times 1 \times 2 \times 3$, and the product would not be unique.

Why it matters: can $6^n$ end in 0?

A number ends in $0$ exactly when $10$ divides it. Since $10 = 2 \times 5$, its prime factorisation must hold a $2$ and a $5$.

Now $6^n = (2 \times 3)^n = 2^n \times 3^n$. Its only primes are $2$ and $3$. By the theorem, there is no other way to break it into primes, so a $5$ can never appear. No power of 6 ends in 0.

A worked example. Can $15^n$ end in $0$ for some natural number $n$?

1. The primes. $15^n = 3^n \times 5^n$.
2. What is missing. A $5$ is there, but there is no $2$.
3. The answer. By the uniqueness of the factorisation, a $2$ can never appear. So $15^n$ never ends in $0$. (It always ends in $5$.)

Practise with new numbers

Try it in Python

This program breaks each number into primes by repeated division, and writes the powers.

Common mistakes

• Calling 1 a prime. It has only one factor. The theorem would fail if $1$ were a prime.
• Stopping at a composite factor. $840 = 8 \times 105$ is true, but it is not the prime factorisation. Keep splitting $8$ and $105$.
• Losing repeated primes. $60 = 2 \times 3 \times 5$ is wrong, since $2 \times 3 \times 5 = 30$. The answer is $60 = 2^2 \times 3 \times 5$.
• Guessing that a number is prime. $91$ and $221$ look prime, but $91 = 7 \times 13$ and $221 = 13 \times 17$.

Remember these

• A prime has exactly two factors. $1$ is neither prime nor composite.
• A factor tree or repeated division gives the prime factorisation. Write it with powers, smallest prime first.
• The Fundamental Theorem of Arithmetic: every composite number is a product of primes in exactly one way, apart from the order.
• A number ends in $0$ only if its primes include both $2$ and $5$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Write $294$ as a product of primes.

2. Is $1$ a prime number? Why does the theorem need the answer?

3. Can $4^n$ end in the digit $0$ for any natural number $n$?

4. A number has the prime factorisation $2^3 \times 3 \times 11$. What is the number?

5. Kabir writes $180 = 2 \times 90$ and says that he has the prime factorisation. Is he right?