Multiples and factors

Every time you share sweets equally, pack eggs into trays or count in twos and fives, you are using mul­ti­ples and fac­tors. They are two of the most use­ful words in the whole of arith­metic. Later you will need them to add frac­tions, to sim­plify frac­tions, and to find the HCF and LCM of num­bers. This les­son builds both ideas slowly from the times tables, so that noth­ing has to be taken on trust.

You know that $3 \times 4 = 12$. This les­son takes that one fact and reads it two ways. Read it one way. It says $12$ is a mul­ti­ple of $3$. Read it the other way. It says $3$ is a fac­tor of $12$. The two words describe the same link between two num­bers. You are just look­ing at it from the other end.

Every­thing below rests on the times tables you have already learned. If a times fact escapes you, do not guess. Go back to the one before it. Then add the num­ber on once more.

Mul­ti­ples

Mul­ti­ply a num­ber by $1$. Then by $2$. Then by $3$. Keep going, on and on. The answers you get are the mul­ti­ples of that num­ber.

Here is the same thing in the acad­e­my's own words. It uses one word you may not know yet. That word is prod­uct. A prod­uct is what a mul­ti­pli­ca­tion gives you. So $12$ is the prod­uct of $3$ and $4$.

The prod­ucts obtained when a num­ber is mul­ti­plied by $1, 2, 3, \ldots$ are called the mul­ti­ples of that num­ber.

That is the rule you have just read, in the acad­e­my's words. You mul­ti­ply the num­ber by $1$, then by $2$, then by $3$. The answers you get are its mul­ti­ples.

So you never have to hunt for mul­ti­ples. You build them. Mul­ti­ply the num­ber by $1$, then by $2$, then by $3$. Go on for as long as you please.

You are mul­ti­ply­ing by each count­ing num­ber in turn. That is the name for $1, 2, 3, 4$ and the rest of the num­bers you count with. Here are the first five mul­ti­ples of $3$. Next to them are the first five of $5$.

Those answers are usu­ally writ­ten short. They go in a list. An equals sign stands in for the word are. This short way of writ­ing is used all through the course.

The short way looks like this: mul­ti­ples of $3 = 3, 6, 9, 12, 15, \ldots$

It means the mul­ti­ples of $3$ are $3, 6, 9, 12, 15$ and on with­out end.

The same short way gives mul­ti­ples of $5 = 5, 10, 15, 20, 25, \ldots$

Look down either col­umn of the grid. Each mul­ti­ple is one whole step big­ger than the one above it. On the left it goes up by three each time. On the right it goes up by five. That is not chance.

Mul­ti­ply­ing is repeated adding. So you get from the fourth mul­ti­ple to the fifth by adding the num­ber on once more.

$3 \times 5$

$$\begin{aligned}3 \times 5 &= (3 \times 4) + 3 \\ &= 12 + 3 \\ &= 15\end{aligned}$$

$5 \times 5$

$$\begin{aligned}5 \times 5 &= (5 \times 4) + 5 \\ &= 20 + 5 \\ &= 25\end{aligned}$$

This is why a list of mul­ti­ples has no end. There is no last count­ing num­ber to mul­ti­ply by. So there can be no last mul­ti­ple. You can always add the num­ber on one more time. The dots at the end of every such list say just that.

Remem­ber. A mul­ti­ple of a num­ber is what you get when you mul­ti­ply it by a count­ing num­ber. A count­ing num­ber is one of $1, 2, 3, 4$ and onwards. Every num­ber is a mul­ti­ple of itself. Why? Mul­ti­ply­ing by $1$ changes noth­ing.

Worked exam­ple: build­ing mul­ti­ples of 7

Sup­pose you want the first six mul­ti­ples of $7$. Start with one lot of seven. Then keep adding seven more.

$$\begin{aligned}7 \times 1 &= 7 \\ 7 \times 2 &= 7 + 7 = 14 \\ 7 \times 3 &= 14 + 7 = 21 \\ 7 \times 4 &= 21 + 7 = 28 \\ 7 \times 5 &= 28 + 7 = 35 \\ 7 \times 6 &= 35 + 7 = 42\end{aligned}$$

So mul­ti­ples of $7 = 7, 14, 21, 28, 35, 42, \ldots$ Each line used only the line before it. You never needed a fact you did not already have.

Worked exam­ple: is a num­ber a mul­ti­ple?

Is $42$ a mul­ti­ple of $6$? Ask whether some count­ing num­ber times $6$ gives $42$. Count up in sixes: $6, 12, 18, 24, 30, 36, 42$. You reach $42$ on the sev­enth step, so $6 \times 7 = 42$. Yes, $42$ is a mul­ti­ple of $6$.

Is $50$ a mul­ti­ple of $6$? Count on: $42, 48, 54$. The count jumps from $48$ straight past $50$ to $54$. It never lands on $50$, so $50$ is not a mul­ti­ple of $6$.

Mul­ti­ples writ­ten out

Read a row across as answers. Then you are read­ing a times table. Read the same row as a list. Then you are read­ing mul­ti­ples. They are the same num­bers wear­ing a dif­fer­ent name.

The dots at the end of each row are doing real work. Either row could have been car­ried on for ever. A row stops only where there was room to stop.

Longer lists of the same kind are writ­ten out in Least Com­mon Mul­ti­ple. There you will find them for $8$, for $9$ and for $12$. They are needed there to find the small­est mul­ti­ple that two or three num­bers share.

Fac­tors

A fac­tor of a num­ber is a num­ber that divides into it exactly. Noth­ing at all is left over.

What is left over after you divide has a name. It is called the remain­der. If the divide comes out exactly, there is no remain­der at all.

Now here is the acad­e­my's own word­ing. In it, "the given num­ber" means the num­ber you started with.

The num­bers which divide the given num­ber with­out remain­der are the fac­tors of the given num­ber.

A fac­tor is also called a divi­sor. The two words mean exactly the same thing. You will meet both.

There is one more word to have ready. Some­times one num­ber divides another exactly. Then we say the sec­ond is divis­i­ble by the first. So $12$ is divis­i­ble by $3$. And $12$ is not divis­i­ble by $5$.

"With­out remain­der" is the whole test. Divide, and look at what is left. If noth­ing at all is left over, the num­ber you divided by is a fac­tor. If some­thing is left over, it is not. Being close does not count.

Take $6$. There is a slow but cer­tain way to find its fac­tors. Try divid­ing it by every num­ber from $1$ up to $6$. Then keep the ones that come out exactly.

Only $4$ and $5$ left any­thing over. The other four divided $6$ exactly. So fac­tors of $6 = 1, 2, 3, 6$.

The same test on $12$ takes twelve lines. It asks exactly the same ques­tion twelve times.

Six of the twelve divi­sions came out exactly. Those six divi­sors are the fac­tors. So fac­tors of $12 = 1, 2, 3, 4, 6, 12$.

Remem­ber. A fac­tor divides its num­ber exactly. If any­thing at all is left over, it is not a fac­tor.

Writ­ing all the fac­tors, in pairs

Later in the course you will meet a method with a long name: 1st method (By writ­ing all fac­tors). The name says what you do first. You write down all the fac­tors.

That is the part to take care over. You must write down every fac­tor. It is easy to miss one.

Look again at the divi­sions above that came out exactly. Take $12 \div 2 = 6$. It says that $2$ divides $12$ with noth­ing over. So $2$ is a fac­tor.

But it says a sec­ond thing at the same time. It says $12$ is two lots of $6$. So $6$ divides $12$ as well. One divi­sion has named two fac­tors.

That is what the pair­ing method uses. Test the num­bers in order: $1$, then $2$, then $3$, and on. Every time one of them divides exactly, write down its part­ner beside it. The part­ner is what the mul­ti­pli­ca­tion gives. $2 \times 6 = 12$, so $2$ and $6$ both go on the list.

Fac­tors of $12$, in pairs

$$\begin{aligned}1 \times 12 &= 12 \\ 2 \times 6 &= 12 \\ 3 \times 4 &= 12\end{aligned}$$

The next num­ber to try is $4$. But $4$ is already writ­ten down as the part­ner of $3$. The two sides have met in the mid­dle. That is the sig­nal to stop.

Any­thing big­ger than $4$ went on the list long ago. It went on the moment its part­ner was found. Read the left col­umn down, then the right col­umn back up. You get fac­tors of $12 = 1, 2, 3, 4, 6, 12$. Those are the same six the twelve-line table found, in three lines.

Remem­ber. To write down every fac­tor, test $1$, $2$, $3$ and on in turn. Put each divi­sor on the list beside its part­ner. So $2 \times 6 = 12$ puts both $2$ and $6$ there. Stop when the num­ber you are about to try is already there as a part­ner. The two sides have met. There is noth­ing left to find.

Three longer lists

Both ways of work­ing will do any num­ber you like. You can divide by every num­ber in turn. Or you can pair. Here are three lists worth hav­ing. Each one is set out as the divi­sions that prove it.

Read each table straight down. The ques­tion is on the left. What it comes to is on the right. Every divi­sion in them comes out exactly, with noth­ing left over. That is what puts each divi­sor on the list.

Fac­tors of $24$

Eight num­bers divide $24$ exactly. $5$ is not one of them. Five goes into $24$ four times and leaves $4$ over. So $24$ is not divis­i­ble by $5$. Nor is $7$, which leaves $3$ over. Nor are $9$, $10$ or $11$.

So fac­tors of $24 = 1, 2, 3, 4, 6, 8, 12, 24$.

Fac­tors of $36$

$36$ has nine fac­tors. Again $5$ is not among them. $36$ is not divis­i­ble by $5$. Five sev­ens are $35$, and one is left over. Notice that $6$ is writ­ten once, although $6 \times 6 = 36$. Here $6$ is its own part­ner. You still put it down one time only. A fac­tor never goes on the list twice.

So fac­tors of $36 = 1, 2, 3, 4, 6, 9, 12, 18, 36$.

Fac­tors of $84$

$84$ has twelve fac­tors. That is more than either num­ber above. $5$ is not among them. Nei­ther are $8$, $9$, $10$ and $11$. Each of those leaves some­thing over.

A list this long is worth build­ing in pairs. Do not trust it to mem­ory. Test $1$, $2$, $3$ and on. Write each divi­sor beside its part­ner.

Fac­tors of $84$, in pairs

$$\begin{aligned}1 \times 84 &= 84 \\ 2 \times 42 &= 84 \\ 3 \times 28 &= 84 \\ 4 \times 21 &= 84 \\ 6 \times 14 &= 84 \\ 7 \times 12 &= 84\end{aligned}$$

None of $5$, $8$, $9$, $10$ and $11$ divides $84$ exactly. So none of them brought a part­ner. The next num­ber to try after $11$ is $12$. But $12$ is already on the list, as the part­ner of $7$. The two sides have met. So the list is com­plete.

The pair­ing set­tles $26$ as well. $26$ is an easy num­ber to write down by mis­take. It is nobody's part­ner here. Twenty-six goes three times into $84$ and leaves $6$ over. But $28$ is the part­ner of $3$, because $3 \times 28 = 84$ exactly.

So fac­tors of $84 = 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84$.

Notice the two columns. The ques­tions run for­wards down the first. The answers run back­wards up the sec­ond. That is the pair­ing, printed as divi­sions. $84 \div 7 = 12$ and $84 \div 12 = 7$ are one fact. So each line puts one fac­tor near the top of the list. Its part­ner goes near the bot­tom.

Fac­tor lists at a glance

Here are seven num­bers with their fac­tors. Read a row across. First comes the num­ber. Then comes every num­ber that divides it exactly, small­est first. Then comes how many there are. The lists for $4$ and $8$ are short enough to check in your head. The rest were worked out above.

Every list begins at $1$. Every list ends at the num­ber itself. Noth­ing out­side those two ends appears any­where. A big­ger num­ber does not always have more fac­tors. Look at $8$ and $6$. $8$ is the big­ger num­ber, but each of them has four fac­tors.

Worked exam­ple: every fac­tor of 48

Test $1$, $2$, $3$ and on. Write each divi­sor beside its part­ner.

$$\begin{aligned}1 \times 48 &= 48 \\ 2 \times 24 &= 48 \\ 3 \times 16 &= 48 \\ 4 \times 12 &= 48 \\ 6 \times 8 &= 48\end{aligned}$$

$5$ leaves $3$ over, so it brings no part­ner. $7$ leaves $6$ over, so it brings none either. The next num­ber to try is $8$, and $8$ is already on the list as the part­ner of $6$. The two sides have met. Read down the left col­umn and back up the right: fac­tors of $48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48$. That is ten fac­tors.

Worked exam­ple: a num­ber with only two fac­tors

Try the same method on $13$. $1 \times 13 = 13$, so $1$ and $13$ go on the list. Now test $2$ and $3$. Each one leaves some­thing over: $1$ for the twos and $1$ for the threes. The next num­ber to try is $4$, but $4 \times 4 = 16$ is already past $13$. Any part­ner of $4$ or of a big­ger num­ber would have to be smaller than $4$, and those have all been tested. So no new pair can appear. Fac­tors of $13 = 1, 13$. A num­ber like this, with exactly two fac­tors, is called a prime num­ber. You will meet primes prop­erly in a later les­son.

Two fac­tors every num­ber has

Every count­ing num­ber divides by $1$ exactly. You get one lot of the num­ber, and noth­ing over. Every count­ing num­ber divides by itself exactly too. You get one of itself, and noth­ing over.

So $1$ and the num­ber itself are fac­tors of every num­ber. They are the small­est and the largest it has. That is why every fac­tor list above begins at $1$ and ends at the num­ber.

So no fac­tor can be big­ger than the num­ber it belongs to. Take $6$. It is not divis­i­ble by $7$. Seven will not go into six even once. So $7$ can­not be a fac­tor of $6$.

Mul­ti­ples run the oppo­site way. No mul­ti­ple of $7$ is smaller than $7$. The small­est one is $7$ itself.

Remem­ber. $1$ is a fac­tor of every num­ber. Every num­ber is a fac­tor of itself. A fac­tor list always starts at $1$ and ends at the num­ber.

The same fact, read two ways

$3 \times 4 = 12$. Because that is true, all of these say one and the same thing. $3$ is a fac­tor of $12$. $4$ is a fac­tor of $12$. $12$ is a mul­ti­ple of $3$. $12$ is a mul­ti­ple of $4$. Divid­ing $12$ by either of them leaves noth­ing over. When­ever you can say one, you can say the oth­ers.

There is one big dif­fer­ence between the two words. It is worth keep­ing in mind. A num­ber has only so many fac­tors. You can write every one of them down. Its mul­ti­ples never run out. You can never write them all down. $12$ has six fac­tors, and no last mul­ti­ple at all.

Remem­ber. A num­ber has only so many fac­tors. You can write every one of them down. None is big­ger than the num­ber itself. Its mul­ti­ples never run out. None of them is smaller than the num­ber itself.

Your turn

Write the first five mul­ti­ples of each

Find every fac­tor of each

True or false, and say how you know

1. $6$ is a fac­tor of $24$.
2. $24$ is a mul­ti­ple of $6$.
3. $9$ is a fac­tor of $12$.
4. $1$ is a fac­tor of $36$.
5. $36$ is a mul­ti­ple of $36$.
6. $7$ is a fac­tor of $84$.
7. $5$ is a fac­tor of $12$.
8. $48$ is a mul­ti­ple of $8$.
9. $24$ is a fac­tor of $6$.
10. $84$ has more fac­tors than $36$.

Com­mon mis­takes

• Mix­ing up the two words. A fac­tor goes into a num­ber; a mul­ti­ple comes out of mul­ti­ply­ing it. $3$ is a fac­tor of $12$, and $12$ is a mul­ti­ple of $3$, never the other way round.
• For­get­ting $1$ or the num­ber itself when list­ing fac­tors. Every fac­tor list starts at $1$ and ends at the num­ber.
• Miss­ing a fac­tor from the mid­dle of a list. Work in pairs, and stop only when the two sides meet.
• Writ­ing a square fac­tor twice. In $36$, the fac­tor $6$ is its own part­ner and goes on the list once.
• Start­ing a list of mul­ti­ples at $0$. The first mul­ti­ple is the num­ber times $1$, which is the num­ber itself.
• Think­ing a big­ger num­ber must have more fac­tors. $8$ and $6$ each have four.

Key terms

Prod­uct

The answer you get when you mul­ti­ply num­bers together.

Count­ing num­ber

One of $1, 2, 3, 4$ and onwards.

Mul­ti­ple

A prod­uct of a num­ber and a count­ing num­ber, such as $15 = 5 \times 3$.

Fac­tor (divi­sor)

A num­ber that divides a given num­ber with no remain­der.

Remain­der

What is left over after a divi­sion.

Divis­i­ble

A num­ber is divis­i­ble by another when the divi­sion leaves no remain­der.

Fac­tor pair

Two fac­tors that mul­ti­ply to give the num­ber, such as $3$ and $4$ for $12$.

Prime num­ber

A num­ber with exactly two fac­tors, $1$ and itself.

Answers

Write the first five mul­ti­ples of each

1. $3, 6, 9, 12, 15$
2. $4, 8, 12, 16, 20$
3. $6, 12, 18, 24, 30$
4. $7, 14, 21, 28, 35$
5. $8, 16, 24, 32, 40$
6. $9, 18, 27, 36, 45$
7. $11, 22, 33, 44, 55$
8. $12, 24, 36, 48, 60$
9. $15, 30, 45, 60, 75$
10. $20, 40, 60, 80, 100$
11. $25, 50, 75, 100, 125$
12. $30, 60, 90, 120, 150$

Find every fac­tor of each

1. $1, 2, 5, 10$
2. $1, 2, 7, 14$
3. $1, 3, 5, 15$
4. $1, 2, 4, 8, 16$
5. $1, 2, 3, 6, 9, 18$
6. $1, 2, 4, 5, 10, 20$
7. $1, 3, 7, 21$
8. $1, 5, 25$
9. $1, 2, 4, 7, 14, 28$
10. $1, 2, 3, 5, 6, 10, 15, 30$
11. $1, 2, 4, 8, 16, 32$
12. $1, 2, 4, 5, 8, 10, 20, 40$

True or false, and say how you know

1. True. $24 \div 6 = 4$ with noth­ing left over.
2. True. $6 \times 4 = 24$; this says the same as ques­tion 1.
3. False. $12 \div 9$ leaves $3$ over.
4. True. $1$ is a fac­tor of every num­ber.
5. True. $36 \times 1 = 36$; every num­ber is a mul­ti­ple of itself.
6. True. $84 \div 7 = 12$ with noth­ing left over.
7. False. $12 \div 5$ leaves $2$ over.
8. True. $8 \times 6 = 48$.
9. False. $24$ is big­ger than $6$, and no fac­tor is big­ger than its num­ber. (It is the other way: $6$ is a fac­tor of $24$.)
10. True. $84$ has twelve fac­tors and $36$ has nine.