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

You know the divisibility rules for 2, 3, 4 and 5. Now come the rules for 6, 7, 8, 9, 10 and 11. Each one swaps a long division for a short sum. You will use them to find factors, to spot prime numbers, to find the H.C.F and L.C.M, and to cancel fractions quickly.

Divisibility rule for 6

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

Six is $2 \times 3$ , so a number with a 2 inside it and a 3 inside it has a 6 inside it. It must pass both tests. Passing only one is not enough.

Example 1

Is 7224 divisible by 6?

Test for 2: the units digit is 4, so yes.
Test for 3: $7 + 2 + 2 + 4 = 15$ , and $15 \div 3 = 5$ , so yes.

It passes both tests, so 7224 is divisible by 6. Check: $7224 = 6 \times 1204$ .

Example 2

Is 124 divisible by 6?

Test for 2: the units digit is 4, so yes.
Test for 3: $1 + 2 + 4 = 7$ , and $7 \div 3$ is 2 with remainder 1, so no.

It is divisible by 2 but not by 3. So 124 is not divisible by 6.

Example 3

Is 310152 divisible by 6? Is 31052?

For 310152: it ends in 2, so it passes the test for 2. Add all the digits: $3 + 1 + 0 + 1 + 5 + 2 = 12$ , which is divisible by 3. So 310152 is divisible by 6. Check: $310152 = 6 \times 51692$ .

For 31052: it ends in 2, so it passes the test for 2. But $3 + 1 + 0 + 5 + 2 = 11$ , and 11 is not divisible by 3. So 31052 is not divisible by 6.

Be careful to add every digit. Leaving one out gives the wrong digit sum and the wrong answer.

Divisibility rule for 7

A natural number is divisible by 7 if the difference between twice the units digit and the remaining part of the number is a multiple of 7, or is 0.

Cover the units digit with your finger. Double it. Take that away from the number still showing.

Example 4

Is 861 divisible by 7?

The units digit is 1, and twice 1 is 2. The remaining part is 86.

$86 - 2 = 84$

And $84 = 7 \times 12$ . So 861 is divisible by 7. Check: $861 = 7 \times 123$ .

Example 5

Is 238 divisible by 7?

Twice 8 is 16. The remaining part is 23.

$23 - 16 = 7$

7 is a multiple of 7, so 238 is divisible by 7. Check: $238 = 7 \times 34$ .

Example 6

Is 724 divisible by 7?

Twice 4 is 8. The remaining part is 72.

$72 - 8 = 64$

64 is not a multiple of 7, because $7 \times 9 = 63$ and $7 \times 10 = 70$ . So 724 is not divisible by 7. Check: $724 = 7 \times 103 + 3$ .

If the number you get is still big, use the rule again on it. For 3094: $309 - 8 = 301$ , then $30 - 2 = 28 = 7 \times 4$ . So 3094 is divisible by 7.

A second test for 7

For a six-digit number, split it into two blocks of three digits, starting from the right. Take the smaller block from the bigger one. If the difference is 0 or a multiple of 7, the number is divisible by 7.

For 342384, the blocks are 342 and 384, and $384 - 342 = 42 = 7 \times 6$ . So 342384 is divisible by 7.

Two panels: 861 split as 86 and 1, giving 86 - 2 = 84 = 7 x 12; and 342384 split into blocks 342 and 384, giving 384 - 342 = 42 = 7 x 6 — The two tests for 7, each worked on its own example.

Divisibility rule for 8

A natural number is divisible by 8 if its last three digits, taken together, are divisible by 8.

This works because a thousand is $8 \times 125$ , so every whole thousand divides by 8.

Example 7

Is 93624 divisible by 8? The last three digits make 624, and $624 = 8 \times 78$ . So yes.

Is 9774 divisible by 8? The last three digits make 774, and $774 = 8 \times 96 + 6$ . So no.

Divisibility rule for 9

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

Example 8

Is 6021 divisible by 9? $6 + 0 + 2 + 1 = 9$ , so yes. Check: $6021 = 9 \times 669$ .

Is 9005 divisible by 9? $9 + 0 + 0 + 5 = 14$ , which is not divisible by 9. So no.

A number divisible by 9 is always divisible by 3 as well. But not the other way round: 7224 has digit sum 15, so it is divisible by 3 but not by 9.

Divisibility rule for 10

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

902050 ends in 0, so it is divisible by 10. 9005 ends in 5, so it is not. A number ending in 0 is also divisible by 2 and by 5.

Divisibility rule for 11

Count the places from the right: the units digit is in the 1st place, the tens digit in the 2nd place, and so on. Add the digits at the odd places. Add the digits at the even places. A natural number is divisible by 11 if the difference between these two sums is 0 or divisible by 11.

Example 9

Is 1221 divisible by 11?

Odd places (1st and 3rd from the right): $1 + 2 = 3$ . Even places (2nd and 4th): $2 + 1 = 3$ . The difference is $3 - 3 = 0$ . So yes. Check: $1221 = 11 \times 111$ .

Example 10

Is 6535 divisible by 11?

Odd places: $5 + 5 = 10$ . Even places: $3 + 6 = 9$ . The difference is $10 - 9 = 1$ , which is not 0 and not divisible by 11. So no.

The digits 6, 5, 3, 5 as tiles labelled 4th even, 3rd odd, 2nd even and 1st odd, with odd places 5 + 5 = 10, even places 3 + 6 = 9 and a gap of 1 — Counting places from the right in 6535. The gap of 1 means 11 does not divide it.

The six rules together

Divisible by	The test
6	passes the test for 2 and the test for 3
7	twice the units digit taken from the rest gives 0 or a multiple of 7
8	the last three digits are divisible by 8
9	the sum of the digits is divisible by 9
10	the units digit is 0
11	the odd-place sum and the even-place sum differ by 0 or a multiple of 11

Common mistakes

Passing only one test for 6. 124 is even but not divisible by 3, so it is not divisible by 6.
Leaving out a digit when adding the digits for 3, 6 or 9.
Adding instead of taking away for 7. Take twice the units digit away from the rest.
Counting places from the left for 11. Always start at the units digit.

Key terms

Remaining part: the number left when the units digit is covered. For 861 it is 86.
Odd places and even places: places counted from the right, starting with the units digit as place 1.

Practice

Is 23502 divisible by 6?
Is 12754 divisible by 6?
Is 917 divisible by 7?
Is 735 divisible by 7?
Is 2401 divisible by 7?
Is 73416 divisible by 8?
Is 9126 divisible by 9?
Is 3094 divisible by 9?
Which of 4500, 4505 and 4550 are divisible by 10?
Is 8932 divisible by 11?
Is 3289 divisible by 11?
Test 9504 for 6, 8, 9 and 11.

Answers

Show answers

Yes. It ends in 2, and $2 + 3 + 5 + 0 + 2 = 12$ is divisible by 3.
No. It is even, but $1 + 2 + 7 + 5 + 4 = 19$ is not divisible by 3.
Yes. $91 - 14 = 77 = 7 \times 11$ . In fact $917 = 7 \times 131$ .
Yes. $73 - 10 = 63 = 7 \times 9$ .
Yes. $240 - 2 = 238$ , then $23 - 16 = 7$ . In fact $2401 = 7 \times 343$ .
Yes. The last three digits make 416, and $416 = 8 \times 52$ .
Yes. $9 + 1 + 2 + 6 = 18$ , which is divisible by 9.
No. $3 + 0 + 9 + 4 = 16$ , which is not divisible by 9.
4500 and 4550, because they end in 0.
Yes. Odd places: $2 + 9 = 11$ . Even places: $3 + 8 = 11$ . The difference is 0.
Yes. Odd places: $9 + 2 = 11$ . Even places: $8 + 3 = 11$ . The difference is 0.
Divisible by all four. It is even and $9 + 5 + 0 + 4 = 18$ , so it passes for 6 and for 9. The last three digits make $504 = 8 \times 63$ . For 11, the odd places give $4 + 5 = 9$ and the even places give $0 + 9 = 9$ , a difference of 0.

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

Divis­i­bil­ity rule for 6#

Exam­ple 1#

Exam­ple 2#

Exam­ple 3#

Divis­i­bil­ity rule for 7#

Exam­ple 4#

Exam­ple 5#

Exam­ple 6#

A sec­ond test for 7#

Divis­i­bil­ity rule for 8#

Exam­ple 7#

Divis­i­bil­ity rule for 9#

Exam­ple 8#

Divis­i­bil­ity rule for 10#

Divis­i­bil­ity rule for 11#

Exam­ple 9#

Exam­ple 10#

The six rules together#

Com­mon mis­takes#

Key terms#

Prac­tice#

Answers#