Divisibility rules for 6, 7, 8, 9, 10 and 11

A fac­tor of a num­ber divides it exactly. Noth­ing is left over. The fac­tors of $6$ are $1, 2, 3$ and $6$. Is $7$ a fac­tor of $342384$? That is the same as ask­ing one thing. Can $342384$ be divided by $7$ with noth­ing left over? You would rather not do that long divi­sion to find out.

A divis­i­bil­ity rule answers the ques­tion with a much smaller sum. You have met the rules for $2$, $3$, $4$ and $5$. This les­son takes the next six. Each one comes with the rea­son it works. A rule you under­stand is a rule you can trust. You can also build it again when you have for­got­ten it.

Every rule below talks about a nat­ural num­ber. The nat­ural num­bers are the count­ing num­bers $1, 2, 3, 4$ and so on. That is every num­ber in this les­son.

You will use these rules far beyond this les­son: to can­cel frac­tions quickly, to find fac­tors and the HCF and LCM, to check a long mul­ti­pli­ca­tion, and to spot at once whether an answer in a test can be right.

The rule for 6

A nat­ural num­ber is divis­i­ble by $6$ if it is divis­i­ble by both $2$ and $3$.

Six is $2 \times 3$. Take a num­ber with a $2$ inside it and a $3$ inside it. That num­ber has a $6$ inside it. Why? Because $2$ and $3$ are coprimes. They share no fac­tor except $1$. So the two tests do not over­lap. Each test finds some­thing the other one can­not.

Watch the two tests fit together in $7224$. It is even. So it is $2$ lots of $3612$. The dig­its of $3612$ add to $12$. So $3612$ is $3$ lots of $1204$. The $2$ and the $3$ are both there. One sits inside the other.

7224

$$\begin{aligned}7224 &= 2 \times 3612 \\ &= 2 \times 3 \times 1204 \\ &= 6 \times 1204\end{aligned}$$

Now $124$. It passes the first test. It fails the sec­ond one. A num­ber only has to fail one test. There is a $2$ in $124$. But there is no $3$ in it at all. So $6$ can­not divide it.

124

$$\begin{aligned}124 &= 2 \times 62 \\ 1 + 2 + 4 &= 7 \\ 124 &= 6 \times 20 + 4\end{aligned}$$

This only works because $2$ and $3$ are coprimes. A pair that is not coprime will not work. Look at $12$. It is divis­i­ble by $2$. It is also divis­i­ble by $4$. But $12$ is not $8$ lots of any­thing: $12 = 8 \times 1 + 4$.

Why not? Because $2$ and $4$ share a fac­tor. The $2$ inside $12$ gets counted twice. So the two tests together only ever show that $4$ is a fac­tor. Noth­ing like that can hap­pen with $2$ and $3$.

Remem­ber. A num­ber is divis­i­ble by $6$ if it passes the test for $2$ and the test for $3$. One test on its own is not enough.

The rule for 7

Take the unit digit of a nat­ural num­ber and dou­ble it. Take that away from the rest of the num­ber. The num­ber is divis­i­ble by $7$ if what is left is a mul­ti­ple of $7$, or is $0$.

Cover the unit digit with a fin­ger. Dou­ble it. Take that from the num­ber still show­ing. Try $861$. Cover the $1$. That leaves $86$. Twice $1$ is $2$.

861

$$\begin{aligned}86 - 2 \times 1 &= 84 \\ 84 &= 7 \times 12 \\ 861 &= 7 \times 123\end{aligned}$$

$84$ is a mul­ti­ple of $7$. So $861$ is a mul­ti­ple of $7$ too.

Some­times the dou­bled digit is the big­ger of the two. Then take the smaller one from the big­ger one. The test asks for the size of the gap. It does not care which way round you write it. Look at $49$. Cover the $9$. That leaves $4$. Twice $9$ is $18$. So the gap is $18 - 4 = 14$. That is $7 \times 2$. And $49 = 7 \times 7$.

Why is cov­er­ing a digit allowed to set­tle it? Dou­bling the unit digit and tak­ing it away is a big­ger move than it looks. It takes away $21$ lots of that digit. And $21$ is a mul­ti­ple of $7$.

Why the dou­bling works

$$\begin{aligned}861 - 21 \times 1 &= 840 \\ 840 &= 10 \times 84\end{aligned}$$

Take a mul­ti­ple of $7$ away from a num­ber. That can­not change whether the num­ber is divis­i­ble by $7$. So $861$ and $840$ stand or fall together. And $840$ is ten lots of $84$. Seven and ten are coprimes. So the ten can­not be the one giv­ing the $7$. If $840$ is a mul­ti­ple of $7$, the $7$ must be in the $84$. That is why the small num­ber decides the big one.

If the num­ber left is still too big to judge, use the rule again on it. Try $3094$. Cover the $4$: $309 - 8 = 301$. Is $301$ a mul­ti­ple of $7$? Use the rule once more: $30 - 2 = 28 = 7 \times 4$. So $301$ is, and so is $3094$. Indeed $3094 = 7 \times 442$.

A sec­ond test

There is a sec­ond test for $7$. You may use whichever one you like. Split the num­ber into two blocks of three dig­its. Start from the right. Take one block from the other. The num­ber is divis­i­ble by $7$ if the gap is $0$ or a mul­ti­ple of $7$.

$342384$ splits into $342$ and $384$.

342384

$$\begin{aligned}384 - 342 &= 42 \\ 42 &= 7 \times 6 \\ 342384 &= 7 \times 48912\end{aligned}$$

The rea­son is $1001$. Look at it closely: $1001 = 7 \times 143$. So one $1001$ is a mul­ti­ple of $7$. Two of them make a mul­ti­ple of $7$ as well. Any whole num­ber of $1001$s is a mul­ti­ple of $7$.

Why the split into threes works

$$\begin{aligned}342384 &= 342 \times 1000 + 384 \\ &= 342 \times 1001 + (384 - 342) \\ &= 342 \times 1001 + 42\end{aligned}$$

The first part is $342$ lots of $1001$. Every one of those is a mul­ti­ple of $7$. That stays true what­ever the front block is. So the whole of that part can be dropped. Only the $42$ is left to test.

Remem­ber. Dou­ble the unit digit and take it from the rest. Or split the num­ber into two blocks of three. Then take one block from the other. Either way you are left with a small num­ber to test.

The rule for 8

A nat­ural num­ber is divis­i­ble by $8$ if the last three dig­its are divis­i­ble by $8$. Read those three dig­its as one num­ber.

One thou­sand is $8 \times 125$. So every whole thou­sand is a mul­ti­ple of $8$. The thou­sands part of a num­ber can never change the answer. Only what is left below the thou­sands can. That part is the last three dig­its.

93624

$$\begin{aligned}93624 &= 93 \times 1000 + 624 \\ 624 &= 8 \times 78 \\ 93624 &= 8 \times 11703\end{aligned}$$

$8$ goes into $624$ exactly, sev­enty-eight times. So $93624$ is divis­i­ble by $8$ as well. Now try $9774$. Its last three dig­its are $774$.

9774

$$\begin{aligned}774 &= 8 \times 96 + 6 \\ 9774 &= 8 \times 1221 + 6\end{aligned}$$

$774$ leaves $6$ over. The bit left over has a name. It is called the remain­der. So $9774$ leaves the same remain­der of $6$. It is not divis­i­ble by $8$. The nine thou­sand added noth­ing to the remain­der, just as promised.

One more: $17352$. The last three dig­its make $352$, and $352 = 8 \times 44$. So $17352$ is divis­i­ble by $8$; in fact $17352 = 8 \times 2169$. If the last three dig­its are still awk­ward, halve them three times. $352$ halves to $176$, then $88$, then $44$. No remain­der appeared, so $8$ divides it.

Remem­ber. Every whole thou­sand is a mul­ti­ple of $8$. So cover every­thing except the last three dig­its.

The rule for 9

A nat­ural num­ber is divis­i­ble by $9$ if the sum of all its dig­its is divis­i­ble by $9$.

Every digit in a num­ber sits in a place. The place tells you what the digit is worth. A digit in the tens place is worth that many tens. A digit in the hun­dreds place is worth that many hun­dreds. Ten, a hun­dred and a thou­sand are the place val­ues.

This rule looks like magic. Then you look at what the place val­ues are doing. $9$, $99$ and $999$ are all mul­ti­ples of $9$. Every place value is one more than one of them. Ten is $9 + 1$. A hun­dred is $99 + 1$. A thou­sand is $999 + 1$.

So each digit hands over a pile of nines. It keeps only itself behind. What is left at the end is the sum of the dig­its.

6021

$$\begin{aligned}6021 &= 6 \times 999 + 0 \times 99 + 2 \times 9 + (6 + 0 + 2 + 1) \\ &= 5994 + 0 + 18 + 9 \\ 6021 &= 9 \times 669\end{aligned}$$

Every­thing before the bracket is a pile of nines. The bracket holds $6 + 0 + 2 + 1 = 9$. That is a mul­ti­ple of $9$. So $6021$ is divis­i­ble by $9$.

9005

$$\begin{aligned}9 + 0 + 0 + 5 &= 14 \\ 9005 &= 9 \times 999 + 14 \\ &= 8991 + 14\end{aligned}$$

Here the left­over is $14$. That is not a mul­ti­ple of $9$. So $9005$ is not divis­i­ble by $9$ either.

The test for $3$ used this very same sum. Any total that is a mul­ti­ple of $9$ is also a mul­ti­ple of $3$. So a num­ber that is divis­i­ble by $9$ is always divis­i­ble by $3$ too. Look: $6021 = 3 \times 2007$. It does not work the other way round. The dig­its of $7224$ add to $15$. That is a mul­ti­ple of $3$, but not of $9$. And $7224$ is divis­i­ble by $3$ and not by $9$.

Remem­ber. Add the dig­its. Is that total divis­i­ble by $9$? Then so is the num­ber. If the total is not, nei­ther is the num­ber.

The rule for 10

A nat­ural num­ber is divis­i­ble by $10$ if its unit digit is $0$.

Ten lots of any num­ber is that num­ber with a $0$ writ­ten after it. That is what mov­ing into the tens col­umn means. So every mul­ti­ple of ten ends in $0$. Noth­ing else does. Any other digit in the units place is a left­over. It is what stayed behind after the tens were taken out.

902050

$$902050 = 90205 \times 10$$

$902050$ ends in $0$. So it is exactly $90205$ lots of ten, with noth­ing over. Now take a num­ber that does not end in $0$.

9005

$$9005 = 900 \times 10 + 5$$

$9005$ has a $5$ in the units place. That $5$ is the remain­der. Now notice one more thing. A num­ber end­ing in $0$ ends in an even digit. So it passes the test for $2$. It also ends in $0$, so it passes the test for $5$. Every mul­ti­ple of ten is a mul­ti­ple of both.

Remem­ber. Only a $0$ in the units place makes a num­ber divis­i­ble by $10$. A $5$ there passes the test for $5$. It does not pass this one.

The rule for 11

Count the places from the right. Add the dig­its at the odd places. Add the dig­its at the even places. Then find the gap between the two sums. A nat­ural num­ber is divis­i­ble by $11$ if that gap is $0$ or is divis­i­ble by $11$.

Count­ing from the right is the part to watch. The units digit sits in the 1st place. That place is odd. The tens digit sits in the 2nd place. That place is even. The places then take turns all the way to the front.

1221

$$\begin{aligned}(1 + 2) - (2 + 1) &= 3 - 3 = 0 \\ 1221 &= 11 \times 111\end{aligned}$$

The two sums are equal. So the gap is $0$, and $0$ counts. $1221$ is divis­i­ble by $11$. Now take $6535$. Its odd places hold $5$ and $5$. Its even places hold $3$ and $6$.

6535

$$\begin{aligned}(5 + 5) - (3 + 6) &= 10 - 9 = 1 \\ 6535 &= 11 \times 594 + 1\end{aligned}$$

The gap is $1$. That is not $0$, and it is not a mul­ti­ple of $11$. So $6535$ is not divis­i­ble by $11$. Now look at what that $1$ turned out to be. It is the remain­der itself. This hap­pens when the odd places make the big­ger sum. The gap must also be smaller than $11$. Then the gap is the remain­der every time. That is not by chance.

Two more before the rea­son. For $2178$, the odd places hold $8$ and $1$, which add to $9$; the even places hold $7$ and $2$, which also add to $9$. The gap is $0$, so $2178 = 11 \times 198$. For $70587$, the odd places give $7 + 5 + 7 = 19$ and the even places give $8 + 0 = 8$. The gap is $11$, a mul­ti­ple of $11$, so $70587 = 11 \times 6417$.

Here is the rea­son. $11$, $99$ and $1001$ are all mul­ti­ples of $11$. Every place value sits right beside one of them.

Take the $3$ in the tens place of $6535$. Ten is one less than $11$. So those three tens are three elevens with $3$ taken off. Those three elevens are a mul­ti­ple of $11$, so they can be ignored. The $3$ can­not be ignored. It still has to be taken away. So a digit in the tens place is taken away, not added.

The $5$ in the hun­dreds place works the other way round. A hun­dred is one more than $99$. So five hun­dreds are five lots of $99$, with $5$ over. And $99$ is $9$ elevens. So five lots of $99$ is five lots of $9$ elevens. That comes to $45$ elevens, which can be ignored. The $5$ left over is added on.

A thou­sand is one less than $1001$. So a digit in the thou­sands place is taken away, just like a tens digit.

Why the two sums work

$$\begin{aligned}6535 &= 6 \times 1001 + 5 \times 99 + 3 \times 11 + [(5 + 5) - (3 + 6)] \\ &= 6006 + 495 + 33 + 1\end{aligned}$$

Add up every­thing left behind. The dig­its at the odd places are added on. The dig­its at the even places are taken away. So you are left with the odd-place total minus the even-place total. That is the very gap the rule asks for.

Some­times the even places make the big­ger sum. Then take the smaller sum from the big­ger one. The test asks for the size of the gap. It does not care which way round you write it.

Remem­ber. Start at the units and add every other digit. Add up the rest. Are the two totals equal? Or is the gap between them a mul­ti­ple of $11$? If either one is true, the num­ber is divis­i­ble by $11$. If nei­ther is true, it is not.

The six rules together

Here are the six rules on one page. Each one comes with the exam­ple it was worked on. Cover the mid­dle col­umn first. See if you can say the test your­self before you look.

Every one of these rules swaps a long divi­sion for a short sum. That is all they are for. Does a rule ever leave you unsure? Then do the divi­sion. The rule was only ever a short cut to the same answer.

Your turn

Which of these are divis­i­ble by 6? Give both tests for each.

Test these for 7 by dou­bling the unit digit

Test these for 7 by split­ting them into two blocks of three

Which of these are divis­i­ble by 8?

Which of these are divis­i­ble by 9?

Which of these are divis­i­ble by 10?

Which of these are divis­i­ble by 11?

Every rule at once

1. Test $7920$ by all six rules in this les­son. Which one does it fail?
2. Test $5544$ by all six rules in this les­son. Which one does it fail?

Com­mon mis­takes

• Using only one test for $6$. $124$ is even but its dig­its add to $7$, so it is not divis­i­ble by $6$.
• Com­bin­ing tests that share a fac­tor, such as $2$ and $4$ for $8$. Only coprime pairs can be com­bined.
• Adding the dou­bled unit digit in the rule for $7$ instead of tak­ing it away.
• Check­ing only the last two dig­its for $8$. The rule needs the last three.
• Mix­ing up the rules for $3$ and $9$. A digit sum of $15$ passes for $3$ but not for $9$.
• Accept­ing a $5$ in the units place for $10$. Only a $0$ passes.
• Count­ing places from the left in the rule for $11$, or for­get­ting that a gap of $0$ counts as a pass.

Key terms

Fac­tor

A num­ber that divides another exactly, leav­ing noth­ing over.

Divis­i­ble

Able to be divided by a num­ber with remain­der $0$.

Divis­i­bil­ity rule

A short test that decides divis­i­bil­ity with­out long divi­sion.

Coprimes

Two num­bers whose only com­mon fac­tor is $1$, such as $2$ and $3$.

Remain­der

What is left over after divid­ing.

Place value

What a digit is worth because of where it sits: ones, tens, hun­dreds and so on.

Digit sum

The total of all the dig­its of a num­ber.

Nat­ural num­ber

A count­ing num­ber $1, 2, 3, \dots$

Answers

Divis­i­ble by 6?

1. $4128$: even, and $4 + 1 + 2 + 8 = 15$ is a mul­ti­ple of $3$. Yes; $4128 = 6 \times 688$.
2. $5130$: even, and $5 + 1 + 3 + 0 = 9$. Yes; $5130 = 6 \times 855$.
3. $2345$: odd, so it fails the test for $2$ (its digit sum $14$ fails for $3$ too). No.
4. $936$: even, and $9 + 3 + 6 = 18$. Yes; $936 = 6 \times 156$.
5. $7215$: its digit sum $15$ passes for $3$, but it is odd. No.
6. $1002$: even, and $1 + 0 + 0 + 2 = 3$. Yes; $1002 = 6 \times 167$.

Divis­i­ble by 7, dou­bling the unit digit

1. $133$: $13 - 6 = 7$. Yes; $133 = 7 \times 19$.
2. $259$: $25 - 18 = 7$. Yes; $259 = 7 \times 37$.
3. $402$: $40 - 4 = 36$, not a mul­ti­ple of $7$. No.
4. $105$: $10 - 10 = 0$. Yes; $105 = 7 \times 15$.
5. $224$: $22 - 8 = 14$. Yes; $224 = 7 \times 32$.
6. $508$: $50 - 16 = 34$, not a mul­ti­ple of $7$. No.

Divis­i­ble by 7, blocks of three

1. $224448$: $448 - 224 = 224 = 7 \times 32$. Yes.
2. $105210$: $210 - 105 = 105 = 7 \times 15$. Yes.
3. $213426$: $426 - 213 = 213$, and $213 = 7 \times 30 + 3$. No.
4. $156702$: $702 - 156 = 546 = 7 \times 78$. Yes.

Divis­i­ble by 8?

1. $51288$: $288 = 8 \times 36$. Yes.
2. $7136$: $136 = 8 \times 17$. Yes.
3. $4506$: $506 = 8 \times 63 + 2$. No.
4. $90032$: $032 = 32 = 8 \times 4$. Yes.
5. $2358$: $358 = 8 \times 44 + 6$. No.
6. $61000$: the last three dig­its are $000$, which is $0$. Yes.

Divis­i­ble by 9?

1. $3618$: digit sum $18$. Yes.
2. $4527$: digit sum $18$. Yes.
3. $8100$: digit sum $9$. Yes.
4. $2345$: digit sum $14$. No.
5. $76518$: digit sum $27$. Yes.
6. $1234$: digit sum $10$. No.

Divis­i­ble by 10?

1. $4570$: ends in $0$. Yes.
2. $3005$: ends in $5$. No.
3. $88000$: ends in $0$. Yes.
4. $10101$: ends in $1$. No.

Divis­i­ble by 11?

1. $2453$: odd places $3 + 4 = 7$, even places $5 + 2 = 7$, gap $0$. Yes.
2. $90816$: odd places $6 + 8 + 9 = 23$, even places $1 + 0 = 1$, gap $22$. Yes.
3. $4829$: odd places $9 + 8 = 17$, even places $2 + 4 = 6$, gap $11$. Yes.
4. $8492$: odd places $2 + 4 = 6$, even places $9 + 8 = 17$, gap $11$. Yes.
5. $7325$: odd places $5 + 3 = 8$, even places $2 + 7 = 9$, gap $1$. No.
6. $1234$: odd places $4 + 2 = 6$, even places $3 + 1 = 4$, gap $2$. No.

Every rule at once

1. $7920$ passes for $6$ (even, digit sum $18$), $8$ ($920 = 8 \times 115$), $9$ (digit sum $18$), $10$ (ends in $0$) and $11$ ($(0 + 9) - (2 + 7) = 0$). It fails the rule for $7$: $792 - 0 = 792 = 7 \times 113 + 1$.
2. $5544$ passes for $6$ (even, digit sum $18$), $7$ ($554 - 8 = 546 = 7 \times 78$), $8$ ($544 = 8 \times 68$), $9$ (digit sum $18$) and $11$ ($(4 + 5) - (4 + 5) = 0$). It fails the rule for $10$, because it ends in $4$.