Prime and composite numbers

What this les­son is about

Every count­ing num­ber can be built by mul­ti­ply­ing. Some num­bers can be built in many ways; $12$ is $2 \times 6$ and also $3 \times 4$. Oth­ers can only be built as one times them­selves. Those are the prime num­bers, and they are the build­ing blocks of all the rest. You will use them when you find the H.C.F and L.C.M, when you can­cel frac­tions, and in later classes when you fac­torise.

Count­ing the fac­tors

You already know what a fac­tor is. A fac­tor of a num­ber divides it exactly. Noth­ing is left over. The fac­tors of $6$ are $1, 2, 3, 6$. The fac­tors of $12$ are $1, 2, 3, 4, 6, 12$. This les­son does one new thing with that idea. It counts them.

$1$ divides every num­ber. And every num­ber divides itself. Those two fac­tors come free. So here is the ques­tion that mat­ters. Does the num­ber have any oth­ers?

That one ques­tion sorts the nat­ural num­bers into three piles. The nat­ural num­bers are the ones you count with. You start at one and keep going for ever. The three piles have names.

Prime num­bers

The acad­e­my's def­i­n­i­tion. A nat­ural num­ber greater than $1$ is a prime num­ber if it has only two divi­sors (fac­tors): $1$ and itself.

Read it slowly. Every part of it does a job. Two divi­sors, not one and not three. $1$ and itself says which two they must be. And greater than $1$ keeps $1$ out. The part called Where $1$ belongs shows why.

Here are five prime num­bers. Their fac­tors are writ­ten out in full.

Every row has exactly two. Look at $7$. Noth­ing else will divide it: not $2$, not $3$, not $4$, not $5$, not $6$. So $7$ keeps just the two fac­tors that every num­ber has. That is what makes a num­ber prime. Not that it is spe­cial. Just that noth­ing else got in.

See­ing it with coun­ters

Here is a way to see the dif­fer­ence. Take some coun­ters and try to lay them out as a rec­tan­gle with at least one full row. Each rec­tan­gle you can make gives you a pair of fac­tors.

Seven coun­ters will only lie in a sin­gle row of seven. Try two rows and one counter is left over. Try three rows and one is left over again. So the only fac­tors are $1$ and $7$. Twelve coun­ters, though, make a row of twelve, two rows of six and three rows of four. Those three rec­tan­gles give the six fac­tors $1, 2, 3, 4, 6, 12$.

Remem­ber. A prime num­ber has exactly two fac­tors: $1$ and itself. Not one fac­tor. Not three. Exactly two.

Com­pos­ite num­bers

Here is the def­i­n­i­tion, the plain way round. A nat­ural num­ber which has more than two divi­sors (fac­tors) is a com­pos­ite num­ber. The acad­emy writes it the other way round. A nat­ural num­ber which is nei­ther $1$ nor prime is called a com­pos­ite num­ber. That says two things. The num­ber is not $1$. And it is not prime. Every other count­ing num­ber is com­pos­ite.

Here are eight com­pos­ite num­bers. Take $4$ first. The fac­tors (divi­sors) of $4$ are $1, 2, 4$. That is three fac­tors. Three is more than two. So $4$ is com­pos­ite.

The same count­ing for the rest of them.

Notice how lit­tle it takes. To show that a num­ber is com­pos­ite, you do not need all of its fac­tors. One extra fac­tor is enough. Then you may stop. Take $9$. It is not even, so $2$ will never divide it. But $3$ does. That third fac­tor set­tles it. The fac­tors of $9$ are $1, 3, 9$. Three fac­tors, and three is more than two.

Remem­ber. A com­pos­ite num­ber has more than two fac­tors. Find one more fac­tor besides $1$ and the num­ber itself. That is enough to prove it.

A worked exam­ple: $24$

To list every fac­tor with­out miss­ing any, work in pairs. Start at $1$ and go up, writ­ing each fac­tor with its part­ner.

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

Five does not divide $24$. The next num­ber to try is $6$, but $6$ is already in the list as the part­ner of $4$. Once the pairs start to repeat, you have found them all. So the fac­tors of $24$ are $1, 2, 3, 4, 6, 8, 12, 24$. That is eight fac­tors, so $24$ is com­pos­ite.

Now try $49$. It is odd, so $2$ does not divide it. Nor do $3$, $4$, $5$ or $6$. But $49 = 7 \times 7$. Here the pair is one num­ber used twice, so it adds just one fac­tor. The fac­tors are $1, 7, 49$: three of them, so $49$ is com­pos­ite.

Where $1$ belongs

$1$ is nei­ther prime nor com­pos­ite. Here is why. The fac­tors of $1$ are $1$. That is one fac­tor, and only one. Look at the two free fac­tors of $1$. The first one is $1$. The sec­ond one is the num­ber itself, which is also $1$. They are the same num­ber. So they fold into one.

So $1$ fails both tests at once. It does not have two fac­tors. So it can­not be prime. It does not have more than two. So it can­not be com­pos­ite. It sits on its own. That is the third pile.

Remem­ber. $1$ is nei­ther prime nor com­pos­ite. It has one fac­tor only. Prime asks for two. Com­pos­ite asks for more than two.

$2$ is the only even prime num­ber

The two def­i­n­i­tions you have just read set­tle some­thing about even num­bers. $2$ divides every even num­ber. So $2$ is a fac­tor of every even num­ber. Now take any even num­ber big­ger than $2$. Its fac­tor list holds $1$. It holds $2$. It holds the num­ber itself. Those are three dif­fer­ent num­bers. Three is more than two. So the num­ber can­not be prime.

The com­pos­ite table above shows it hap­pen­ing. The fac­tors of $4$ are $1, 2, 4$. The fac­tors of $6$ are $1, 2, 3, 6$. The fac­tors of $8$ are $1, 2, 4, 8$. The same $2$ caught all three. That is why each one has more than two fac­tors.

$2$ is the one even num­ber this argu­ment can­not touch. Look at what hap­pens with $2$. The free fac­tor the num­ber itself is $2$. The fac­tor $2$ is that same num­ber. So noth­ing new is added. The list stays at $1, 2$. That is exactly two fac­tors. So $2$ is prime. Every other even num­ber is ruled out. So every prime after $2$ is odd.

Remem­ber. $2$ is the only even prime num­ber. Every even num­ber big­ger than $2$ has $1$, $2$ and itself among its fac­tors. That is more than two. So no other even num­ber can be prime.

Four things to remem­ber

1. The small­est prime num­ber is $2$. $1$ is not prime. So $2$ is the first count­ing num­ber with exactly two fac­tors.
2. The small­est com­pos­ite num­ber is $4$. $1$ is nei­ther. $2$ and $3$ are prime. So $4$ is the first num­ber to pick up a third fac­tor.
3. $1$ is nei­ther prime nor com­pos­ite.
4. There are $25$ prime num­bers below $100$.

Here is the list. Read it across each row.

Twenty-five alto­gether. Count them ten at a time. That is a good way to check you have them all. It also shows how unevenly they fall.

There are four in the first ten. There is only one in the last. But the count does not fall steadily on the way. The for­ties hold three. The sev­en­ties hold three too. Noth­ing you have learned so far tells you where the next prime will fall. You learn the list.

Prov­ing that a num­ber is com­pos­ite

One extra fac­tor is all it takes. You have just learned the divis­i­bil­ity rules. Each rule is a quick test. It tells you whether some small num­ber divides your num­ber. You do not have to do the divid­ing sum. Those tests are made for find­ing a fac­tor. Try $91$. It looks like a prime. It is not.

Start with the easy tests. $91$ ends in $1$. The tests for $2$, for $5$ and for $10$ all look at the last digit. A last digit of $1$ passes none of them. So $2$ does not divide $91$. Nor does $5$. Nor does $10$.

Next add the dig­its: $9 + 1 = 10$. The tests for $3$ and for $9$ use that total. But $3$ does not divide $10$. So $3$ does not divide $91$ either. Nor does $9$.

The test for $6$ asks for two things at once. Both $2$ and $3$ must divide the num­ber. $91$ has just failed both. So $6$ does not divide it.

The test for $4$ looks at the last two dig­its. The test for $8$ looks at the last three. But $91$ has only two dig­its. So both tests look at the whole of $91$. And $4$ does not divide $91$. Nor does $8$.

That leaves $7$ and $11$ to try.

Now try the rule for $7$. You met it in Divis­i­bil­ity rules for 6, 7, 8, 9, 10 and 11. Here is what it says. Take the last digit and dou­ble it. Then take that away from the num­ber the other dig­its make. Look at what you are left with. Is it $0$? Is it in the $7$ times table? Then $7$ divides the num­ber.

Test­ing $91$ with the rule for $7$

$$9 - 2 \times 1 = 9 - 2 = 7$$

$7$ is in the $7$ times table. So $7$ divides $91$. Divide it out. Now you can see what $91$ was hid­ing.

What $91$ really is

$$91 = 7 \times 13$$

So the fac­tors of $91$ are $1, 7, 13, 91$. That is four fac­tors. So $91$ is com­pos­ite. That is why it is miss­ing from the list of twenty-five.

The other way round is harder. It is worth being hon­est about that. To show a num­ber is com­pos­ite, you find one fac­tor. To show a num­ber is prime, you need much more. You must know that no extra fac­tor exists at all. Fail­ing to find one is not the same as there being none. That is why the twenty-five primes below $100$ are worth learn­ing by heart. Know­ing them is quicker than work­ing them out again every time.

Remem­ber. One extra fac­tor proves a num­ber com­pos­ite. Prov­ing a num­ber prime is a dif­fer­ent kind of job. So learn the twenty-five.

A sec­ond worked exam­ple: $119$

This num­ber is above $100$, so the list of twenty-five can­not help. Work through the tests. It ends in $9$, so $2$, $5$ and $10$ do not divide it. Its dig­its add to $1 + 1 + 9 = 11$, which $3$ does not divide, so nei­ther $3$ nor $9$ divides it.

Now the rule for $7$. Dou­ble the last digit and take it from the rest.

$$11 - 2 \times 9 = 11 - 18 = -7$$

The result is $-7$. The rule works with a neg­a­tive answer too: $-7$ is $7$ times $-1$. So $7$ divides $119$, and $119 = 7 \times 17$. The fac­tors are $1, 7, 17, 119$, so $119$ is com­pos­ite.

The first twenty num­bers, sorted

Every­thing in this les­son, gath­ered in one place. The table holds every num­ber from $1$ to $20$. It gives the fac­tors of each one. It gives how many there are. It gives which pile that puts it in. Noth­ing here is new.

Read the last col­umn down­wards. $1$ stands alone at the top. The primes are scat­tered through the rest. They fol­low no pat­tern you could pre­dict. Every­thing else is com­pos­ite. Now read the mid­dle col­umn. You can see just why each num­ber landed where it did.

Your turn

Write out all the fac­tors of each num­ber, then say whether it is prime or com­pos­ite

Which of these are prime?

And these, in words

1. Why is $1$ nei­ther prime nor com­pos­ite?
2. How many prime num­bers are there below $100$? Write them all out from mem­ory. Then check your list against the one above.
3. $87$ is not a prime num­ber. Find the fac­tor that proves it. Which rule found it for you?
4. Write down the small­est prime num­ber. Say why no smaller num­ber will do. Now write down the small­est com­pos­ite num­ber. Say why no smaller num­ber will do there either.
5. A num­ber has exactly three fac­tors. Is it prime or com­pos­ite? How do you know?
6. Between $90$ and $100$ there is only one prime num­ber. Which one is it?

Com­mon mis­takes

• Call­ing $1$ a prime num­ber. It has only one fac­tor, and a prime needs exactly two.
• Think­ing every odd num­ber is prime. $9$, $15$, $21$, $49$ and $91$ are all odd and all com­pos­ite.
• Think­ing $2$ can­not be prime because it is even. It has exactly two fac­tors, $1$ and $2$.
• Stop­ping too soon when test­ing. Not being divis­i­ble by $2$, $3$ and $5$ does not make a num­ber prime; $49$ and $91$ are caught by $7$.
• Count­ing a repeated fac­tor twice. The fac­tors of $25$ are $1, 5, 25$: three, not four.

Key terms

Nat­ural num­bers

The count­ing num­bers $1, 2, 3$ and so on.

Fac­tor (divi­sor)

A num­ber that divides another num­ber exactly, leav­ing noth­ing over.

Prime num­ber

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

Com­pos­ite num­ber

A nat­ural num­ber with more than two fac­tors.

Divis­i­bil­ity rule

A quick test that shows whether one num­ber divides another with­out doing the divi­sion.

Fac­tor pair

Two fac­tors that mul­ti­ply to give the num­ber, such as $3$ and $8$ for $24$.

Answers

Fac­tors, then prime or com­pos­ite

1. $21$: fac­tors $1, 3, 7, 21$. Com­pos­ite.
2. $23$: fac­tors $1, 23$. Prime.
3. $25$: fac­tors $1, 5, 25$. Com­pos­ite.
4. $27$: fac­tors $1, 3, 9, 27$. Com­pos­ite.
5. $29$: fac­tors $1, 29$. Prime.
6. $31$: fac­tors $1, 31$. Prime.
7. $33$: fac­tors $1, 3, 11, 33$. Com­pos­ite.
8. $35$: fac­tors $1, 5, 7, 35$. Com­pos­ite.
9. $37$: fac­tors $1, 37$. Prime.
10. $39$: fac­tors $1, 3, 13, 39$. Com­pos­ite.

Which of these are prime?

1. $41$: prime.
2. $45$: not prime, since $45 = 5 \times 9$.
3. $49$: not prime, since $49 = 7 \times 7$.
4. $51$: not prime, since $51 = 3 \times 17$.
5. $53$: prime.
6. $57$: not prime, since $57 = 3 \times 19$.
7. $59$: prime.
8. $63$: not prime, since $63 = 7 \times 9$.
9. $67$: prime.
10. $69$: not prime, since $69 = 3 \times 23$.

So the primes are $41$, $53$, $59$ and $67$.

In words (model answers)

1. $1$ has only one fac­tor, itself. A prime needs exactly two fac­tors and a com­pos­ite needs more than two, so $1$ is nei­ther.
2. There are $25$: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97$.
3. The dig­its of $87$ add to $8 + 7 = 15$, which $3$ divides. So the rule for $3$ finds the fac­tor $3$, and $87 = 3 \times 29$.
4. The small­est prime is $2$: the only smaller nat­ural num­ber is $1$, which has one fac­tor. The small­est com­pos­ite is $4$: $1$ is nei­ther, and $2$ and $3$ are prime.
5. Com­pos­ite. It has more than two fac­tors, and a prime has exactly two. (Such num­bers are squares of primes, like $4$, $9$ and $25$.)
6. $97$.