Prime factorisation and the standard form of a number

Think of prime num­bers as the build­ing blocks of all whole num­bers, the way let­ters are the build­ing blocks of words. Every com­pos­ite num­ber is built by mul­ti­ply­ing primes together. Break­ing a num­ber back into its primes is called prime fac­tori­sa­tion. You will use it to find the H.C.F and L.C.M, to sim­plify frac­tions, and later to work with square roots and alge­bra.

You know how to list the fac­tors of a num­ber. You know which num­bers are prime. This les­son joins the two ideas. Every com­pos­ite num­ber can be taken apart into primes. There is only one way to do it.

Look at $24$. Its fac­tors are $1, 2, 3, 4, 6, 8, 12$ and $24$. Only two of them are prime: $2$ and $3$. So if you keep split­ting $24$ until noth­ing will split any more, $2$s and $3$s are what you are left hold­ing. All the other fac­tors are com­pos­ite, except $1$. And $1$ is nei­ther prime nor com­pos­ite.

What prime fac­tori­sa­tion means

Two words are needed first. Every num­ber divides by $1$. Every num­ber divides by itself as well. A num­ber is com­pos­ite when some other num­ber divides it too. Take $24$. It divides by $2$, and $2$ is nei­ther $1$ nor $24$. So $24$ is com­pos­ite.

The other word is prod­uct. A prod­uct is what you get when you mul­ti­ply num­bers together. So $2 \times 2 \times 2 \times 3$ is a prod­uct. It is the prod­uct of three $2$s and one $3$.

Take any com­pos­ite num­ber $N$. You can write it as a prod­uct of prime fac­tors. There is one and only one way to do it. That prod­uct is called the prime fac­tori­sa­tion of $N$. To fac­torise a num­ber means to take it apart like this.

That rule promises two things. The first is that you can always fin­ish. This part is easy to see.

Start with a com­pos­ite num­ber. Some smaller fac­tor divides it. Take that fac­tor out. What comes away is prime or com­pos­ite. If it is prime, that piece is done. If it is com­pos­ite, it splits again.

Every split leaves smaller num­bers than before. Every piece is a whole num­ber, and no piece is smaller than $2$. So the pieces can­not keep get­ting smaller for ever. Below $2$ there is nowhere left to go. The split­ting has to stop, and it can only stop when every piece is prime.

The sec­ond promise is that you always fin­ish with the same primes. That one is shown fur­ther down, once $24$ has been taken apart.

Tak­ing 24 apart

Divide by the small­est prime that goes into the num­ber. Then divide the answer the same way. Keep going until you reach $1$.

Prime fac­tori­sa­tion of 24

$$\begin{aligned}24 \div 2 &= 12 \\ 12 \div 2 &= 6 \\ 6 \div 2 &= 3 \\ 3 \div 3 &= 1 \\ 24 &= 2 \times 2 \times 2 \times 3\end{aligned}$$

$24$ is even, so $2$ divides it. The $12$ it leaves is even too. So is the $6$ after that. That brings you to $3$. And $3$ is prime. Noth­ing but $3$ itself will divide it.

The last divi­sion gives $1$. That is where you stop. You divided by $2, 2, 2$ and $3$ along the way. Those are the prime fac­tors.

The $1$ at the bot­tom is never writ­ten among the fac­tors. Mul­ti­ply­ing by $1$ changes noth­ing. And $1$ is nei­ther prime nor com­pos­ite.

Only one answer

The rule promises there is only one answer. $24$ can show you what that means. You can break it open at $4 \times 6$. Or you can break it open at $3 \times 8$. Keep going either way. See where you end up.

24 bro­ken open two ways

$$\begin{aligned}24 &= 4 \times 6 \\ &= (2 \times 2) \times (2 \times 3) \\ &= 2 \times 2 \times 2 \times 3 \\ 24 &= 3 \times 8 \\ &= 3 \times (2 \times 2 \times 2) \\ &= 2 \times 2 \times 2 \times 3\end{aligned}$$

One route began at $4 \times 6$. The other began at $3 \times 8$. Both fin­ished with three $2$s and one $3$. The lad­der above gave that same answer. Say two peo­ple fac­torise the same num­ber. If they end with dif­fer­ent primes, one of them has slipped.

Draw­ing it as a fac­tor tree

A fac­tor tree is a pic­ture of the same split­ting. Write the num­ber at the top. Draw two branches down to any two fac­tors whose prod­uct is that num­ber. Any branch that ends in a prime stops there; ring it. Any branch that ends in a com­pos­ite num­ber splits again.

The two trees have dif­fer­ent shapes, but the green cir­cles at the ends hold exactly the same primes. That is the promise of one and only one way, seen in a pic­ture.

Remem­ber. A com­pos­ite num­ber can be writ­ten as a prod­uct of prime fac­tors. There is one and only one way to do it.

Stan­dard form

Three $2$s mul­ti­plied together give $2 \times 2 \times 2 = 8$. You write that as $2^{3}$. The base $2$ says what is being mul­ti­plied. The small raised num­ber is the index. It counts how many times the base was used.

A power writ­ten in this short way, like $2^{3}$, is called the expo­nen­tial form. You met pow­ers and the expo­nen­tial form in Let­ters that stand for num­bers. That short way of writ­ing is all this les­son needs.

Start with a num­ber you have taken apart. You found $24 = 2 \times 2 \times 2 \times 3$. Count the three $2$s with an index. That gives $24 = 2^{3} \times 3$. There is a prime, and above it a count. Then another prime, and its count. That is all stan­dard form is.

Now write the rule out for any num­ber at all. The count­ing num­bers $1, 2, 3, \ldots$ are the nat­ural num­bers. Take a nat­ural num­ber $N$ big­ger than $1$.

The rule has to fit every num­ber, so it uses let­ters in place of the primes. The small num­bers below the let­ters count noth­ing. They are only labels. $p_{1}$ means the first prime. $p_{2}$ means the sec­ond. And $k$ stands for how many dif­fer­ent primes the num­ber has.

The small num­bers above the let­ters are the indices. Those ones do count. Dots in the mid­dle mean the list car­ries on in the same way.

One word in the rule needs care. It is dis­tinct. It means each prime is writ­ten down once only. Say a prime turns up three times. You do not write it three times. You write it once, with an index of $3$. A prime that does not divide the num­ber is left out.

Now here is the rule itself. The stan­dard form of $N$ is $N = p_{1}^{a_{1}} \times p_{2}^{a_{2}} \times \cdots \times p_{k}^{a_{k}}$. The let­ters $p_{1}, p_{2}, \ldots, p_{k}$ are dis­tinct primes. The let­ters $a_{1}, a_{2}, \ldots, a_{k}$ are nat­ural num­bers. Each one is at least $1$.

Stan­dard form of 54

$54$ ends in $4$, so it is even. That means $2$ divides it. It leaves $27$, which is odd. So $2$ is fin­ished with.

The dig­its of $27$ add to $2 + 7 = 9$. And $3$ goes into that total. So $3$ goes into $27$ as well. That is the next prime to divide by. It goes in three times.

Stan­dard form of 54

$$\begin{aligned}54 \div 2 &= 27 \\ 27 \div 3 &= 9 \\ 9 \div 3 &= 3 \\ 3 \div 3 &= 1 \\ 54 &= 2 \times 3 \times 3 \times 3 \\ &= 2^{1} \times 3^{3}\end{aligned}$$

An index of $1$ tells you noth­ing new. So it is not usu­ally writ­ten. The stan­dard form is left as $54 = 2 \times 3^{3}$.

When a prime is miss­ing: 100

Not every prime turns up in every num­ber. $100$ is even, and so is the $50$ it leaves. So two $2$s come out. That leaves $25$.

It is odd, so $2$ is done. Its dig­its add to $2 + 5 = 7$. And $3$ does not go into that total. So no $3$ will come out of $25$. But $25$ ends in $5$. So $5$ divides it.

Stan­dard form of 100

$$\begin{aligned}100 \div 2 &= 50 \\ 50 \div 2 &= 25 \\ 25 \div 5 &= 5 \\ 5 \div 5 &= 1 \\ 100 &= 2 \times 2 \times 5 \times 5 \\ &= 2^{2} \times 5^{2}\end{aligned}$$

There is no $3$ in $100$. There is no $7$ either. Those primes are just miss­ing. A miss­ing prime is not writ­ten down. Stan­dard form records the primes a num­ber is made of. It records noth­ing else.

Remem­ber. In stan­dard form each prime is writ­ten once. An index says how many times it was used.

A big­ger num­ber: 360

Larger num­bers take more steps, but the method is exactly the same. Keep divid­ing by $2$ while you can, then by $3$, then by $5$, and so on.

Stan­dard form of 360

$$\begin{aligned}360 \div 2 &= 180 \\ 180 \div 2 &= 90 \\ 90 \div 2 &= 45 \\ 45 \div 3 &= 15 \\ 15 \div 3 &= 5 \\ 5 \div 5 &= 1 \\ 360 &= 2 \times 2 \times 2 \times 3 \times 3 \times 5 \\ &= 2^{3} \times 3^{2} \times 5\end{aligned}$$

One with a larger prime: 84

$$\begin{aligned}84 \div 2 &= 42 \\ 42 \div 2 &= 21 \\ 21 \div 3 &= 7 \\ 7 \div 7 &= 1 \\ 84 &= 2^{2} \times 3 \times 7\end{aligned}$$

After $21 \div 3$ you are left with $7$. Nei­ther $2$, $3$ nor $5$ divides it, and $7$ is prime, so it divides itself and you reach $1$.

When to stop try­ing primes

Sup­pose what is left is a num­ber like $97$. Try $2, 3, 5, 7$ in turn. None divides it. The next prime is $11$, and $11 \times 11 = 121$ is already big­ger than $97$. If $97$ had a fac­tor big­ger than $11$, the other fac­tor in the pair would be smaller than $11$, and you have tried all of those. So $97$ is prime. Once a prime mul­ti­plied by itself passes the num­ber, you can stop.

Why one and only one way mat­ters

A num­ber's prime fac­tori­sa­tion is a name that belongs to it alone. $2 \times 2 \times 2 \times 3$ is $24$ and can be noth­ing else. And $24$ is $2 \times 2 \times 2 \times 3$ and can be noth­ing else.

Look at $24 = 2^{3} \times 3$ and $54 = 2 \times 3^{3}$. They are built from the very same two primes. What tells them apart is the indices. Change how many times each prime is used. You get a dif­fer­ent num­ber.

Now set the two lists side by side. You have $24 = 2 \times 2 \times 2 \times 3$ and $54 = 2 \times 3 \times 3 \times 3$. At a glance you can see what they share. Both have a $2$ and a $3$. You can see what they do not share too.

That works because each num­ber has one list and no other. Later on, this is how num­bers get com­pared. H.C.F is short for high­est com­mon fac­tor. L.C.M is short for least com­mon mul­ti­ple. Both come in later lessons. The acad­emy calls this way of find­ing them the 2nd method (Prime fac­tor­iza­tion method).

Remem­ber. Each num­ber has one list of primes and no other. So the list can stand in for the num­ber itself.

The three num­bers of this les­son

The mid­dle col­umn writes every prime out in full. Each prime is there as many times as the num­ber needs. The right-hand col­umn says the same thing more shortly. An index counts the repeats.

Quick tests that save time

• $2$ divides a num­ber that ends in $0, 2, 4, 6$ or $8$.
• $3$ divides a num­ber whose dig­its add to a mul­ti­ple of $3$. For $135$, $1 + 3 + 5 = 9$.
• $5$ divides a num­ber that ends in $0$ or $5$.
• For $7$, $11$ and larger primes, just try the divi­sion.

Your turn

Write each num­ber as a prod­uct of primes

Write each of these in stan­dard form

Check every answer by mul­ti­ply­ing your primes back together. You should get the num­ber you started with. If you do not, divide again one prime at a time. The mis­take will show itself.

Com­mon mis­takes

• Stop­ping at a com­pos­ite fac­tor. $24 = 2 \times 2 \times 6$ is not fin­ished, because $6$ is not prime.
• Writ­ing $1$ as a prime fac­tor. $1$ is nei­ther prime nor com­pos­ite.
• Get­ting the index wrong. $2 \times 2 \times 2$ is $2^{3}$, not $2 \times 3$.
• Writ­ing the same prime twice in stan­dard form, as in $2^{2} \times 2$. Com­bine it: $2^{3}$.
• Not check­ing. Mul­ti­ply the primes back together and make sure you get the num­ber you started with.

Key terms

Prime num­ber

A num­ber greater than $1$ whose only fac­tors are $1$ and itself, such as $2, 3, 5, 7$.

Com­pos­ite num­ber

A num­ber greater than $1$ that has some fac­tor other than $1$ and itself.

Prod­uct

The answer when num­bers are mul­ti­plied together.

Prime fac­tori­sa­tion

Writ­ing a num­ber as a prod­uct of primes, in one and only one way.

Fac­tor tree

A branch­ing pic­ture that splits a num­ber into fac­tors until every branch ends in a prime.

Index

The small raised num­ber that counts how many times the base is mul­ti­plied, as the $3$ in $2^{3}$.

Stan­dard form

A prime fac­tori­sa­tion writ­ten with each dis­tinct prime once and an index, as in $360 = 2^{3} \times 3^{2} \times 5$.

Answers

Write each num­ber as a prod­uct of primes

1. $20 = 2 \times 2 \times 5$
2. $28 = 2 \times 2 \times 7$
3. $30 = 2 \times 3 \times 5$
4. $40 = 2 \times 2 \times 2 \times 5$
5. $44 = 2 \times 2 \times 11$
6. $50 = 2 \times 5 \times 5$
7. $56 = 2 \times 2 \times 2 \times 7$
8. $63 = 3 \times 3 \times 7$
9. $66 = 2 \times 3 \times 11$
10. $75 = 3 \times 5 \times 5$
11. $96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3$
12. $98 = 2 \times 7 \times 7$
13. $99 = 3 \times 3 \times 11$
14. $135 = 3 \times 3 \times 3 \times 5$
15. $189 = 3 \times 3 \times 3 \times 7$
16. $200 = 2 \times 2 \times 2 \times 5 \times 5$

Write each of these in stan­dard form

1. $90 = 2 \times 3^{2} \times 5$
2. $120 = 2^{3} \times 3 \times 5$
3. $144 = 2^{4} \times 3^{2}$
4. $180 = 2^{2} \times 3^{2} \times 5$
5. $250 = 2 \times 5^{3}$
6. $360 = 2^{3} \times 3^{2} \times 5$
7. $625 = 5^{4}$
8. $800 = 2^{5} \times 5^{2}$
9. $1000 = 2^{3} \times 5^{3}$
10. $1500 = 2^{2} \times 3 \times 5^{3}$