Divisibility rules for 2, 3, 4 and 5

Sup­pose you have $126$ sweets to share equally among some chil­dren, or $452$ chairs to set out in equal rows. Before doing any long divi­sion, it helps to know whether the shar­ing will come out exact. Divis­i­bil­ity rules answer that ques­tion in a few sec­onds, just by look­ing at the dig­its.

You will use these rules again and again: when you find fac­tors, when you sim­plify frac­tions, when you find the H.C.F. and L.C.M. of num­bers, and when you check your own mul­ti­pli­ca­tion. They are worth know­ing well, and worth under­stand­ing, because a rule you under­stand is a rule you will not mix up.

One num­ber divides another when it goes in exactly. Noth­ing is left over.

Take $6$ and $12$. $6$ divides $12$, because $12 \div 6 = 2$ with no remain­der. Now take $13$. $6$ does not divide $13$. Here $13 \div 6$ is $2$, with $1$ left over.

When one num­ber divides another exactly, we say the sec­ond is divis­i­ble by the first.

You know this idea by another name. The num­bers that divide a num­ber exactly are its fac­tors. Mul­ti­ples and fac­tors sets them out in full.

You could test any num­ber by divid­ing it. Then you look at what is left over. But for $2$, $3$, $4$ and $5$ you do not have to. Each one has a rule of its own. The rule reads the answer off the dig­its. It takes a moment.

This les­son gives you the four rules. It gives them the way the acad­emy states them. It also shows why each rule is true.

The rules are for nat­ural num­bers. Those are the count­ing num­bers $1, 2, 3, 4$ and on and on. You use them to count things.

Where the rules come from

Every num­ber is built out of place value. That means each digit has a place of its own. The place tells you what the digit is worth.

Look at $452$. The $4$ sits in the hun­dreds place. It means four hun­dreds. The $5$ sits in the tens place. It means five tens. The $2$ means two units. So $452 = 400 + 50 + 2$.

Now look at what a ten is made of. A ten is $2 \times 5$. So any whole num­ber of tens is divis­i­ble by $2$. It is divis­i­ble by $5$ too.

Now look at a hun­dred. A hun­dred is $4 \times 25$. So any whole num­ber of hun­dreds is divis­i­ble by $4$.

All that divid­ing is done before you even look at the num­ber. Only the part left behind can decide the answer. That part is the last digit. Or it is the last two dig­its. The rule for $3$ works in another way. Its own part of the les­son shows how.

Divis­i­bil­ity Rule for 2

A nat­ural num­ber is divis­i­ble by $2$ if the digit at the unit place of the num­ber is either $0$ or $2$ or $4$ or $6$ or $8$.

That is the rule. Here is how you use it. Look at the digit at the unit place. Is it $0$ or $2$ or $4$ or $6$ or $8$? Then the num­ber is divis­i­ble by $2$.

The unit place is the last digit. It is the one on the right. Split the num­ber there. Then you can see why one digit is enough.

$452$ is $450$ and $2$. The $450$ is forty-five tens. Every ten is $2 \times 5$. So $450$ can be halved exactly. The count of tens makes no dif­fer­ence. That leaves only the $2$ at the end. And $2$ halves cleanly into $2 \times 1$.

You can also see this with objects. Put $452$ but­tons into pairs. The $45$ bags of ten pair off per­fectly, five pairs to a bag. Only the $2$ loose but­tons are left to check, and they make one more pair.

Is 452 divis­i­ble by 2?

$$\begin{aligned}452 &= 450 + 2 \\ &= 2 \times 225 + 2 \times 1 \\ &= 2 \times (225 + 1) \\ &= 2 \times 226\end{aligned}$$

$953$ splits the same way. It splits into $950$ and $3$. The $950$ is ninety-five tens. It gives no trou­ble at all. But $3$ is $2 \times 1$ with $1$ still over. So the whole num­ber leaves a remain­der of $1$. That is why $953$ is not divis­i­ble by $2$.

Is 953 divis­i­ble by 2?

$$\begin{aligned}953 &= 950 + 3 \\ &= 2 \times 475 + 2 \times 1 + 1 \\ &= 2 \times 476 + 1\end{aligned}$$

This is the same thing as even and odd. An even num­ber leaves $0$ when you divide by $2$. An odd num­ber leaves $1$. Nat­ural num­bers, whole num­bers, even and odd sets that out.

So this rule takes the even num­bers. It turns down the odd ones. $452$ is even. $953$ is odd. The unit digit told you so. You did no divid­ing at all.

Remem­ber. For $2$, only the unit digit mat­ters. Every­thing above it is a whole num­ber of tens. And every ten is already even.

Divis­i­bil­ity Rule for 3

A nat­ural num­ber is divis­i­ble by $3$ if the sum of all its dig­its is divis­i­ble by $3$.

Here is how you use it. Add up all the dig­its. Is that sum divis­i­ble by $3$? Then the num­ber is divis­i­ble by $3$ too.

This rule looks odder than the oth­ers. Here is why. Ten does not divide by $3$ exactly. It leaves $1$ over.

But look at ten again. Ten is one more than $9$. A hun­dred is one more than $99$. A thou­sand is one more than $999$. And $9$, $99$ and $999$ all divide by $3$ exactly.

So a hun­dred splits into two parts. One part is $99$, which divides by $3$. The other part is the $1$ left over.

A ten splits in the same way. One part is $9$, which divides by $3$. The other part is $1$ again. A unit is just $1$.

Now make two heaps. Put every piece that divides by $3$ into the first heap. Put every left­over $1$ into the sec­ond heap.

The sec­ond heap holds one $1$ for each hun­dred. It holds one $1$ for each ten. It holds one $1$ for each unit. So the sec­ond heap is the sum of the dig­its. The num­ber divides by $3$ when that heap does. It does not when the heap does not.

Is 117 divis­i­ble by 3?

$$\begin{aligned}117 &= 100 + 10 + 7 \\ &= (99 + 1) + (9 + 1) + 7 \\ &= (99 + 9) + (1 + 1 + 7) \\ &= 108 + 9 \\ &= 3 \times 36 + 3 \times 3 \\ &= 3 \times 39\end{aligned}$$

The $108$ is the first heap. It holds the $99$ from the hun­dred. It holds the $9$ from the ten. And $108$ is $3 \times 36$.

The $9$ is the sec­ond heap. That $9$ is the digit sum $1 + 1 + 7$. And it is $3 \times 3$. Both heaps divide by $3$. So $117$ is divis­i­ble by $3$.

$7874$ fails this test. It fails by just as much as its digit sum does. The dig­its add to $7 + 8 + 7 + 4 = 26$. That is $3 \times 8$ with $2$ over. The num­ber itself is $3 \times 2624$ with the same $2$ over.

The digit sum of 7874

$$\begin{aligned}7 + 8 + 7 + 4 &= 26 \\ &= 3 \times 8 + 2\end{aligned}$$

And 7874 itself

$$7874 = 3 \times 2624 + 2$$

A big­ger one: is 4782 divis­i­ble by 3?

Add the dig­its. If the sum is still large, you may add its dig­its again.

$$\begin{aligned}4 + 7 + 8 + 2 &= 21 \\ 2 + 1 &= 3\end{aligned}$$

The digit sum $21$ is $3 \times 7$, so $4782$ is divis­i­ble by $3$. Divid­ing con­firms it: $4782 = 3 \times 1594$.

There is a short­cut while you add. Any digit that is $0$, $3$, $6$ or $9$ already divides by $3$, so you may leave it out of the sum. In $9360$ every digit, $9$, $3$, $6$ and $0$, can be dropped, so noth­ing is left over at all. So $9360$ is divis­i­ble by $3$, and indeed $9360 = 3 \times 3120$.

Remem­ber. For $3$, add the dig­its. A num­ber and its digit sum leave the same remain­der. That is the remain­der you get when you divide by $3$. So if one of them is divis­i­ble by $3$, the other must be too.

Divis­i­bil­ity Rule for 4

A nat­ural num­ber is divis­i­ble by $4$ if the last two dig­its of the num­ber, all together, are divis­i­ble by $4$.

Here is how you use it. Take the last two dig­its, all together. Are they divis­i­ble by $4$? Then the num­ber is divis­i­ble by $4$.

A hun­dred is $4 \times 25$. So a whole num­ber of hun­dreds always divides by $4$. Cut the num­ber after the hun­dreds. Every­thing you cut off has passed the test already. Only the last two dig­its are left to check.

Is 76532 divis­i­ble by 4?

$$\begin{aligned}76532 &= 76500 + 32 \\ &= 765 \times 4 \times 25 + 4 \times 8 \\ &= 4 \times 19133\end{aligned}$$

Start with the $76500$. Count it in hun­dreds. There are seven hun­dred and sixty-five of them. Every hun­dred is $4 \times 25$. So this part divides by $4$.

Now take the last two dig­its. They are $32$. That is $4 \times 8$. Both parts divide by $4$. So the whole num­ber does too: $76532 \div 4 = 19133$.

Is 126 divis­i­ble by 4?

$$\begin{aligned}126 &= 100 + 26 \\ &= 4 \times 25 + 4 \times 6 + 2 \\ &= 4 \times 31 + 2\end{aligned}$$

The hun­dred is no trou­ble. But $26$ is $4 \times 6$ with $2$ over. So $126$ is $4 \times 31$ with the same $2$ over.

Now look at $126$ again. It ends in $6$. So it passes the test for $2$. But pass­ing for $2$ does not set­tle the ques­tion for $4$.

Two more to try: 3500 and 7318

$3500$ ends in $00$. The num­ber made by the last two dig­its is $0$, and $0$ divides by $4$ (it is $4 \times 0$). So $3500$ is divis­i­ble by $4$: $3500 = 4 \times 875$.

$7318$ ends in $18$, and $18 = 4 \times 4 + 2$. So $7318$ leaves the same remain­der, $2$: $7318 = 4 \times 1829 + 2$.

If the last two dig­its are hard to judge, halve them twice. $76$ halves to $38$, and $38$ halves to $19$. Both halv­ings worked, so $76$ divides by $4$. But $18$ halves to $9$, and $9$ is odd, so $18$ does not.

Remem­ber. For $4$, cover every­thing but the last two dig­its. What you cov­ered is a whole num­ber of hun­dreds. And every hun­dred is $4 \times 25$.

Divis­i­bil­ity Rule for 5

A nat­ural num­ber is divis­i­ble by $5$ if the digit at the unit place of the num­ber is either $0$ or $5$.

Here is how you use it. Look at the digit at the unit place again. Is it $0$ or $5$? Then the num­ber is divis­i­ble by $5$.

The rea­son is the one that gave you the rule for $2$. Read it the other way round. A ten is $5 \times 2$. So any whole num­ber of tens divides by $5$ exactly.

Only the unit digit is left to check. There are ten dig­its it could be. Out of those ten, only $0$ and $5$ divide by $5$ with no remain­der.

Is 5785 divis­i­ble by 5?

$$\begin{aligned}5785 &= 5780 + 5 \\ &= 578 \times 5 \times 2 + 5 \\ &= 5 \times 1157\end{aligned}$$

Is 6021 divis­i­ble by 5?

$$\begin{aligned}6021 &= 6020 + 1 \\ &= 602 \times 5 \times 2 + 1 \\ &= 5 \times 1204 + 1\end{aligned}$$

Count $6020$ in tens. There are six hun­dred and two of them. So $6020$ divides by $5$ exactly. But the $1$ on the end does not. That one unit is the whole remain­der.

Think of coins. Only ₹5 coins and ₹10 notes are allowed. Any amount you can pay exactly with them ends in $0$ or $5$. An amount like ₹6021 can­not be paid exactly: one rupee is always left over.

Remem­ber. Two and five are part­ners, because $2 \times 5 = 10$. That is why both rules look only at the unit digit.

The four rules side by side

Here are the four rules together. The rea­son for each one sits beside it. You have met all these rea­sons already. The table is only a place to look them up.

Pass­ing one rule does not set­tle the next one. Take $7874$. It ends in $4$, so it is divis­i­ble by $2$. But its dig­its add to $26$, so it is not divis­i­ble by $3$.

Now take $6021$. Its dig­its add to $9$, so it is divis­i­ble by $3$. But it ends in $1$, so it is not divis­i­ble by $5$.

One link does run between two of them. Say a num­ber divides by $4$. Then you can split it into fours. Each four is two twos. So you can split the num­ber into twos as well. That means it divides by $2$. Any­thing the rule for $4$ takes, the rule for $2$ takes too.

Look at $76532$. It is divis­i­ble by $2$, as its last digit shows. The link does not run back­wards. $126$ is divis­i­ble by $2$. It still fails the test for $4$.

Check the rules by divid­ing

A rule is worth noth­ing if the divi­sion does not agree with it. Here are the four num­bers that passed. Each one is divided out in full.

Every one is exact. Noth­ing is left over any­where. Now do the same with the four that failed. Try $953$ by $2$, $7874$ by $3$, $126$ by $4$ and $6021$ by $5$. You will find some­thing left over each time.

Your turn

Which of these are divis­i­ble by 2?

Which of these are divis­i­ble by 3? Write down the digit sum you used.

Which of these are divis­i­ble by 4? Write down the last two dig­its you used.

Which of these are divis­i­ble by 5?

Test each of these against all four rules. Write down which of 2, 3, 4 and 5 divide it

1. $120$
2. $135$
3. $244$
4. $350$
5. $936$
6. $1005$

Com­mon mis­takes

• Using the last digit for 3. $13$ ends in $3$ but is not divis­i­ble by $3$; its digit sum is $4$. The rule for $3$ always uses all the dig­its.
• Using only the last digit for 4. $126$ ends in $6$, which is even, but $26$ does not divide by $4$. You need the last two dig­its together.
• Adding the last two dig­its for 4. For $316$ you test the num­ber $16$, not $1 + 6 = 7$.
• Think­ing the link runs back­wards. Every num­ber divis­i­ble by $4$ is divis­i­ble by $2$, but not every even num­ber is divis­i­ble by $4$.
• For­get­ting 0 as a unit digit. $460$ and $90000$ end in $0$, so they are divis­i­ble by both $2$ and $5$.

Key terms

Divides

Goes into a num­ber exactly, with noth­ing left over.

Divis­i­ble

Able to be divided exactly by a given num­ber: $12$ is divis­i­ble by $6$.

Remain­der

What is left over after divid­ing as far as pos­si­ble.

Fac­tor

A num­ber that divides another exactly. $6$ is a fac­tor of $12$.

Unit digit

The last digit of a num­ber, in the units (ones) place.

Digit sum

The total of all the dig­its of a num­ber. The digit sum of $117$ is $9$.

Nat­ural num­bers

The count­ing num­bers $1, 2, 3, 4$ and so on.

Answers

Which of these are divis­i­ble by 2?

1. $138$: yes, it ends in $8$.
2. $275$: no, it ends in $5$.
3. $460$: yes, it ends in $0$.
4. $1234$: yes, it ends in $4$.
5. $9087$: no, it ends in $7$.
6. $50006$: yes, it ends in $6$.

Which of these are divis­i­ble by 3?

1. $141$: digit sum $6$, yes.
2. $232$: digit sum $7$, no.
3. $501$: digit sum $6$, yes.
4. $7002$: digit sum $9$, yes.
5. $8888$: digit sum $32$, no.
6. $12345$: digit sum $15$, yes.

Which of these are divis­i­ble by 4?

1. $316$: last two dig­its $16$, yes.
2. $522$: last two dig­its $22$, no.
3. $1128$: last two dig­its $28$, yes.
4. $4530$: last two dig­its $30$, no.
5. $76540$: last two dig­its $40$, yes.
6. $9012$: last two dig­its $12$, yes.

Which of these are divis­i­ble by 5?

1. $705$: yes.
2. $832$: no.
3. $1460$: yes.
4. $2225$: yes.
5. $30001$: no.
6. $90000$: yes.

Test each of these against all four rules

1. $120$: divis­i­ble by $2$, $3$, $4$ and $5$ (ends in $0$, digit sum $3$, last two dig­its $20$).
2. $135$: divis­i­ble by $3$ and $5$ only (odd, digit sum $9$, ends in $5$).
3. $244$: divis­i­ble by $2$ and $4$ only (digit sum $10$, last two dig­its $44$).
4. $350$: divis­i­ble by $2$ and $5$ only (digit sum $8$, last two dig­its $50$ not divis­i­ble by $4$).
5. $936$: divis­i­ble by $2$, $3$ and $4$ (digit sum $18$, last two dig­its $36$), not by $5$.
6. $1005$: divis­i­ble by $3$ and $5$ only (odd, digit sum $6$, last two dig­its $05$).