Divisibility Rules for 2, 3, 4, 5, 6, 9 and 10

Take a number like 6,351. To find out whether 3 divides it you could do the long division, or you could add up the digits and know in a second. Every shortcut like that comes out of place value.

Problem

Five claim tags have numbers 2,470; 3,805; 6,118; 9,640; and 7,333. Without dividing, decide which are multiples of 5. Which single digit settles the question? How many of the five tags win a prize?

Show a hint

You do not have to divide the whole tag by 5. Write one tag as a stack of tens plus its leftover ones digit, like 2,470 = 247 tens and 0 ones, and watch which part you can stop worrying about.
A whole number of tens is already a multiple of 5, since every ten is 5 twice. By closure from the last lesson, throwing away a multiple of 5 leaves the question resting on the single ones digit. Which ones digits make the leftover a multiple of 5?
Sort the tags by their last digit alone. A tag wins for 5 when it ends in 0 or 5, and it also wins for 10 only when it ends in 0.

Show the full solution

Read the last digit of each tag.

2, 470

ends in 0,

3, 805

ends in 5, and

9, 640

ends in 0, so those three are multiples of 5, while

6, 118

and

7, 333

end in 8 and 3. That is

3

tags. The ones digit decides it alone, because everything above it is a stack of tens and every ten already holds a 5.

Problem

Band sections have 84,317; 2,000,006; 55,550; and 9,999 players. Which counts are even and can line up in perfect pairs? How many of the four can pair up perfectly?

Show a hint

Pairing up two to a row works exactly when the count is even, so the real question is which of these four counts is even. Look hard at the end of each number before you reach for anything bigger.
Try writing a number as its tens part plus its last digit, like $55, 550 = 55, 55 \times 10 + 0$ or $84, 317 = 8, 431 \times 10 + 7$ . Any whole number of tens is already even, so the only thing still in doubt is that final leftover digit.
For $2, 000, 006$ , all those zeros sit in the tens part and are already even on their own. Only the final $6$ is left to decide the whole question, and $6$ is even.

Show the full solution

Pairing up with nobody left over is just asking whether the count is even, so read the last digit.

84, 317

ends in 7 and

9, 999

ends in 9, both odd.

2, 000, 006

ends in 6 and

55, 550

ends in 0, both even, so

2

sections pair up perfectly. Everything above the units digit is a whole number of tens, and tens are even, so only that last digit can tip a number to odd.

The last digit settled 2, 5, and 10 because every higher place is a multiple of ten. Now try 4. Ten is not a multiple of 4, so one digit is not enough. We need to cut earlier and keep a bigger tail. Where should the cut go?

Problem

A dealer splits stacks of 5,328; 7,914; 13,500; and 60,742 cards into complete hands of 4. Write each as hundreds plus its last two digits. How many stacks deal out evenly?

Show a hint

Try splitting each deck at the hundreds, like $7, 914 = 79 \times 100 + 14$ . One hundred is $25$ full hands of $4$ , so every full hundred already deals out evenly. That means the hundreds part never causes trouble, no matter how big it is.
If the hundreds part always deals out evenly, then whether the whole stack works comes down to that small leftover number in the last two digits. So only check whether $28$ , $14$ , $00$ , and $42$ each split into hands of $4$ .

Show the full solution

One hundred is

4 \times 25

, so every full hundred deals out evenly and only the last two digits matter.

5, 328 = 53 \times 100 + 28, 7, 914 = 79 \times 100 + 14

13, 500 = 135 \times 100 + 00, 60, 742 = 607 \times 100 + 42

The tail

28 = 4 \times 7

works and the empty tail

00

works, while

14

and

42

are not multiples of 4. That leaves

2

stacks. Ten is not a multiple of 4, which is why one digit is not enough here and you have to keep two.

Problem

Arcade machines show totals 47,216; 9,360; and 128,500. Test only the last three digits of each for divisibility by 8. How many are divisible by 8?

Show a hint

A whole number of strips means the total is a multiple of $8$ . Look at $1000$ . Can you write $1000$ itself as a whole number times $8$ ? If every full thousand is already a stack of strips, then the thousands part can be thrown away and only what is left below a thousand can decide things.
Split each total right after the last three digits, like $47, 216 = 47 \times 1000 + 216$ . Closure lets you drop the $47 \times 1000$ chunk because it is a pile of complete strips, so only the three-digit tail, $216$ , is left to test against $8$ .
Now line the tests up. For $2$ you read $1$ digit, for $4$ you read $2$ digits, and for $8$ you read $3$ . Ask which power of ten each of $2, 4, 8$ first divides into cleanly, and the digit counts will line up by themselves.

Show the full solution

One thousand is

8 \times 125

, so every full thousand is a whole number of 8s and only the last three digits matter. For

47, 216

the tail

216 = 8 \times 27

fits, for

9, 360

the tail

360 = 8 \times 45

fits, and for

128, 500

the tail

500

lands between

8 \times 62 = 496

and

8 \times 64 = 512

, so it fails. That is

2

totals. You read as many digits as the first power of ten your divisor divides cleanly, so

2 ∣ 10

4 ∣ 100

, and

8 ∣ 1000

give 1 digit, 2 digits, and 3 digits.

Try the same tail trick on 9. It would need a power of ten that 9 divides, and there is none, since $10 = 9 + 1$ , $100 = 99 + 1$ , and $1000 = 999 + 1$ . Every power of ten misses a multiple of 9 by exactly 1, and because the miss is the same size every time, we can still build a test out of it.

Problem

The stamp number is 5,742. Break it by place value, write each power of ten as a multiple of 9 plus 1, and let the multiples fall away. What leftover decides whether 9 divides 5,742?

Show a hint

Do not divide the whole 5,742 at once. Split it by place value, $5742 = 5 \times 1000 + 7 \times 100 + 4 \times 10 + 2$ , and look at each power of ten on its own.
Write each power of ten as a multiple of 9 plus 1, so $1000 = 999 + 1$ , $100 = 99 + 1$ , $10 = 9 + 1$ . Every ' $+ 1$ ' hands its digit straight back to you.
After the $999$ , $99$ , and $9$ chunks fall away as multiples of 9, the only piece left to check is $5 + 7 + 4 + 2$ .

Show the full solution

Spread

5742

out by place value and write each power of ten as a multiple of 9 plus 1.

5742 = 5 (999 + 1) + 7 (99 + 1) + 4 (9 + 1) + 2 = (5 \times 999 + 7 \times 99 + 4 \times 9) + (5 + 7 + 4 + 2)

The first group is built from

999

99

, and

9

, so closure drops it as a multiple of 9. What decides the question is the digit sum

5 + 7 + 4 + 2 = 18

. Since

18 = 9 \times 2

, the stamp number is divisible by 9, and

5742 \div 9 = 638

confirms it.

Here is the whole idea in one picture. Every power of ten is one short step above a run of nines, since

10 = 9 + 1

100 = 99 + 1

, and

1000 = 999 + 1

, and a run of nines is always a multiple of

9

. So when you write a number as its digits times those powers of ten, each digit splits into two pieces, a chunk of nines that the digit rides on and the bare digit left over. The nine chunks all gather into one big multiple of

9

, which closure lets us set aside, and what stays behind is nothing but the digits added up. That leftover is the digit sum, which is why for

4, 152

you only ever have to check

4 + 1 + 5 + 2 = 12

. The number is divisible by

9

exactly when its digit sum is, because the digit sum is the one piece the multiple of

9

never carried away.

Problem

A view counter reads 5,839,206. Sum its digits, and if the result has more than one digit, sum again. What single digit do you land on, and what does it say about divisibility by 3 and by 9?

Show a hint

Add the seven digits of 5,839,206 together. You should get a two-digit number, so do not stop there.
Your first total is 33. Now add ITS digits, 3 + 3, to crush it down to one digit.
The single digit you reach is the remainder when 5,839,206 is divided by 9. Compare that digit to 3 and to 9 to read off which one divides the count.

Show the full solution

Add the digits,

5 + 8 + 3 + 9 + 2 + 0 + 6 = 33

. That is still two digits, so add again,

3 + 3 = 6

. A number and its digit sum leave the same remainder under 9, so this single digit is that remainder. It is a multiple of 3 but not of 9, so

5, 839, 206

is divisible by 3 and not by 9.

Try 11 next. We have $10 = 11 - 1$ , so ten leaves remainder $- 1$ . Then $100 = 99 + 1$ leaves $+ 1$ , and $1000 = 1001 - 1$ leaves $- 1$ . The remainders flip between $- 1$ and $+ 1$ as you climb the places, so the digits will need alternating signs.

Problem

Cards read 90,728 and 53,481. Exactly one is a multiple of 11. Since

10 \equiv - 1 (m o d 11)

, combine digits with alternating signs. Which card is the multiple?

Show a hint

Start at $10 = 11 - 1$ . What does the next place value, $100$ , sit near? Try writing $100$ as a multiple of $11$ plus a little leftover, and then do the same for $1000$ , watching the leftover flip sign each time.
Since the leftovers go $+ 1, - 1, + 1, - 1, \dots$ from the units place upward, a number leaves the same remainder against $11$ as its digits combined with alternating signs, starting $+$ at the units. Build that alternating sum for each card.
For $90, 728$ the alternating sum from the units is $8 - 2 + 7 - 0 + 9$ . Work that out, do the same for $53, 481$ , and check which result is itself a multiple of $11$ .

Show the full solution

Since

10 = 11 - 1

100 = 9 \times 11 + 1

, and

1000 = 91 \times 11 - 1

, the place values carry signs

+, -, +, -

from the units up once closure throws the multiples of 11 away. Combine each card's digits with those signs.

8 - 2 + 7 - 0 + 9 = 22, 1 - 8 + 4 - 3 + 5 = - 1

Only

22 = 2 \times 11

is a multiple of 11, so the card is

90, 728

. The alternating sum gives the remainder against 11, and

53, 481

misses by exactly that

- 1

So far we can test 2, 3, 4, 5, 8, 9, 10, and 11, but not 6, 12, or 15. Each of those factors into divisors we already know how to test, so it is tempting to run both piece tests and combine the results. That does work, but only for certain choices of the two pieces.

Problem

A clerk thinks that if a pile is even and divisible by 6, then it must be divisible by 12. Every pile below 50 that is divisible by 2 and by 6 but NOT by 12 breaks that rule. Find the sum of all of them.

Show a hint

Being even is no extra news once a pile is divisible by 6, since 6 already carries a factor of 2 inside it. So the piles that break the clerk's rule are the multiples of 6 that are not multiples of 12.
Write each pile as $6 k$ and run $k = 1, 2, 3, \dots$ until the pile passes 50. Then test each of those piles for divisibility by 12, and keep the ones that fail.

Show the full solution

The piles below 50 that break the rule are

6

18

30

and

42

, so the sum is

6 + 18 + 30 + 42 = 96 .

Every multiple of 6 is already even, so the test for 2 adds nothing. Write a pile as

6 k

. Divisibility by 12 needs a second factor of 2, and that only appears when

k

is even, so the rule breaks exactly at odd

k

. The split

12 = 2 \times 6

counts the same factor of 2 twice, while

12 = 2^{2} \times 3

needs two of them. The honest split is

12 = 3 \times 4

Problem

A print shop has 25,830 flyers. Using the matching shortcut for each of 2, 3, 4, 5, and 9, decide which stack sizes divide 25,830 evenly. How many of the five work?

Show a hint

Each divisor has its own window into the number. For 2 and 5 you only need the last digit, for 4 the last two digits, and for 3 and 9 the digit sum. Look at the right part of 25,830 for each test instead of dividing the whole thing.
The flyer count ends in 0, which settles 2, 5, and 4 in one glance, and the digits add to a tidy round total that settles 3 and 9 together. Add 2+5+8+3+0 and see what it is a multiple of.
Four of the five fit. The only one to double check is 4, where ending in 0 is not enough on its own, so peek at the full last two digits, 30, and ask if that is a multiple of 4.

Show the full solution

The last digit is

0

, so 2 and 5 both divide the run. For 4, read the last two digits, and

30 = 4 \times 7 + 2

is not a multiple of 4, so 4 fails. For 3 and 9, read the digit sum.

2 + 5 + 8 + 3 + 0 = 18 = 9 \times 2

So 9 divides the run, and 3 does too since 3 divides every 9. That is

4

of the five stack sizes. Ending in 0 settles 2 and 5 but never 4, which needs the full last two digits.

Run the tests backward. Instead of judging a finished number, suppose one digit is hidden and the number must be divisible by something. The same place-value thinking applies, but now you solve for the missing digit rather than just checking.

Problem

A boarding pass prints the flight number as 46,N12, but the hundreds digit N is smudged. The airline only issues numbers divisible by 9. Use the digit-sum test backward to find N.

Show a hint

The digit-sum test says a number is divisible by $9$ exactly when the sum of its digits is a multiple of $9$ . Add up the digits you can see and leave $N$ sitting in the sum.
You should land on $13 + N$ . Now ask which single digit you can drop in so that $13 + N$ becomes a multiple of $9$ . The next multiple of $9$ past $13$ is $18$ .
Only one digit from $0$ through $9$ makes $13 + N$ reach a multiple of $9$ . Which one closes the gap up to $18$ ?

Show the full solution

Add the digits of

46, N 12

and keep

N

in the sum.

4 + 6 + N + 1 + 2 = 13 + N

For divisibility by 9 this has to be a multiple of 9, and the only one a single digit can reach from 13 is 18. So

13 + N = 18

and

N = 5

. The flight number is

46, 512

, whose digits sum to

18 = 2 \times 9

. The digit-sum test runs backward just as well as forward, which is often enough to recover a hidden digit.

Problem

A label reads 543N6 with the tens digit N smudged. It must pass both the sorter for multiples of 9 and the sorter for multiples of 4. What digit is N?

Show a hint

Two machines means two tests, and each test watches a different part of the number. The multiple of 9 test cares about the whole digit sum, while the multiple of 4 test cares only about the last two digits, N6.
Add the known digits, 5+4+3+6 = 18, so the digit sum is 18+N. For that to be a multiple of 9, N has just two choices. Now check which of those choices makes the last two digits a multiple of 4.

Show the full solution

For the 9 sorter, the digit sum is

5 + 4 + 3 + N + 6 = 18 + N

, which is a multiple of 9 only for

N = 0

(sum 18) or

N = 9

(sum 27). For the 4 sorter, only the last two digits matter, since 100 is a multiple of 4. With

N = 0

the tail is

06 = 6

, not a multiple of 4, but with

N = 9

the tail is

96 = 4 \times 24

. Only one digit clears both machines,

N = 9 .

Two tests read different pieces of the same number, so stacking them pins the hidden digit where the survivor lists overlap.

Every test worked the same way. Cut where one chunk becomes a multiple of the divisor, use closure to drop that chunk, and check the small piece left over. The cut lands in a different place for each divisor. The next section uses these tests to find prime factors.

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Practice

Five raffle tickets have numbers 3,160; 8,002; 47,225; 90,118; and 6,500. Which are divisible by 5? How many win a prize?

Show the solution

Read the last digit of each ticket.

3, 160

ends in 0,

47, 225

ends in 5, and

6, 500

ends in 0, so those three are divisible by 5, while

8, 002

and

90, 118

end in 2 and 8. That is

3

winners. Everything above the units digit is a multiple of ten, and ten is built from a 5, so only the last digit decides.

Practice

Five flight log codes are 1,316; 5,550; 23,128; 8,914; and 40,016. Which are divisible by 4? How many qualify?

Show the solution

One hundred is

4 \times 25

, so every full hundred is a multiple of 4 and closure lets you keep only the last two digits. For

1, 316

the tail is

16 = 4 \times 4

, for

23, 128

it is

28 = 4 \times 7

, and for

40, 016

it is

16

again, all passing. The tails

50

and

14

each sit two past a multiple of 4, so those fail. That leaves

3

codes.

Practice

A track code reads 49,d06. To keep it divisible by 3, what is the largest digit d can be?

Show the solution

The digit sum is

4 + 9 + d + 0 + 6 = 19 + d

, and divisibility by 3 needs that total to be a multiple of 3. Since

19

leaves a remainder of 1,

d

has to supply a remainder of 2, which happens for

d = 2, 5, 8

. The largest single digit there is

8

. With

d = 8

the sum is

27 = 3 \times 9

, while

d = 9

would give 28 and miss.

Practice

Batch codes are 5,431; 7,002; 12,345; 81,000; and 6,453. How many are divisible by 9?

Show the solution

Add the digits of each code, since a number and its digit sum leave the same remainder under 9. The five sums are

5 + 4 + 3 + 1 = 13

7 + 0 + 0 + 2 = 9

1 + 2 + 3 + 4 + 5 = 15

8 + 1 + 0 + 0 + 0 = 9

, and

6 + 4 + 5 + 3 = 18

. The multiples of 9 are

9

9

, and

18

, belonging to

7, 002

81, 000

, and

6, 453

, so

3

codes qualify. The digit sum works because every power of ten is one more than a run of nines.

Practice

Every divisibility test reads only the piece the tossed chunk cannot hide. Write a number as digits times powers of ten, cut where one chunk becomes a multiple of the divisor, and closure throws that chunk away. The last digit settles 2, 5, 10, since the whole upper part is a multiple of all three. The last two digits settle 4 and the last three settle 8, since

100 = 4 \times 25

and

1000 = 8 \times 125

. The digit sum settles 3 and 9, since each power of ten is one above a multiple of 9. The alternating digit sum settles 11, since the powers of ten land just below then above a multiple of 11, flipping sign. A composite settles by splitting it into coprime factors and testing each.

Practice

Five sticker orders are 4,128; 7,533; 90; 25,164; and 3,400. Which are divisible by 6? How many can the printer fill?

Show the solution

A number is divisible by 6 exactly when it is divisible by both 2 and 3.

4, 128

is even with digit sum 15,

90

is even with digit sum 9, and

25, 164

is even with digit sum 18, so those three pass both tests.

7, 533

is odd, and

3, 400

is even but its digit sum 7 is not a multiple of 3. The printer can fill

3

orders. The split

6 = 2 \times 3

works because 2 and 3 share no common factor.

Practice

A bib number reads 34_7 with the tens digit hidden. Every bib is divisible by 9. Use the digit-sum test backward to find the hidden digit.

Show the solution

Call the hidden tens digit

d

. The digit sum is

3 + 4 + d + 7 = 14 + d

, and it must be a multiple of 9. Since

d

runs from 0 to 9, the sum runs from 14 to 23, and the only multiple of 9 in that window is 18, so

14 + d = 18

gives

d = 4

. That makes the bib

3447

, and

9 \times 383 = 3447

. The digit-sum test runs in reverse, turning a hidden digit into a small equation.

Practice

Daily krill counts are 3,128; 17,400; 9,990; 52,016; and 6,300. Which are divisible by 8? How many work?

Show the solution

A thousand is

8 \times 125

, so the thousands part is always a multiple of 8 and only the last three digits matter. For

3, 128

the tail is

128 = 8 \times 16

, for

17, 400

it is

400 = 8 \times 50

, and for

52, 016

it is

016 = 8 \times 2

, so those three work. For

9, 990

the tail

990

sits 6 past

8 \times 123 = 984

, and for

6, 300

the tail

300

sits 4 past

8 \times 37 = 296

, so both fail. That is

3

counts.

Practice

Among house numbers 1 through 500, how many end in 6 and are divisible by 4?

Show the solution

Divisibility by 4 reads only the last two digits, since

100 = 4 \times 25

sweeps every full hundred away. Among the endings

06, 16, 26, 36, 46, 56, 66, 76, 86, 96

, the multiples of 4 are

16, 36, 56, 76, 96

, five per hundred. From 1 through 500 there are 5 full hundreds, so the total is

5 \times 5 = 25

. The winners are exactly the endings with an odd tens digit, which is why the same five repeat in every hundred.

Practice

A count reads 8_,386 with the thousands digit missing. It is divisible by 9. Find the hidden digit, then decide whether the full number is also divisible by 4.

Show the solution

The visible digits add to

8 + 3 + 8 + 6 = 25

, so with the hidden digit

d

the digit sum is

25 + d

. That runs from 25 to 34, and the only multiple of 9 in range is 27, so

d = 2

and the count is

82, 386

. For 4, only the last two digits matter, and

86 = 4 \times 21 + 2

, so

82, 386

is not divisible by 4. Each test reads a different chunk, so passing one never drags the other along.

Practice

A stamp serial reads 7_,320 with the thousands digit smudged. It must be divisible by both 3 and 8. How many digits could hide under the smudge?

Show the solution

For 8, only the last three digits matter, and

320 = 8 \times 40

, so the serial is divisible by 8 whatever hides in the thousands place. For 3, the digit sum is

7 + □ + 3 + 2 + 0 = 12 + □

, and since 12 is already a multiple of 3, the sum works exactly when

□

is a multiple of 3, giving

0, 3, 6, 9

. That is

4

digits. The smudge sits above the window the 8 test reads, so only the 3 test rules anything out.

Practice

Tickets read 2,051; 4,613; 5,000; 8,267; and 3,002. Unlike 2 or 5, the number 7 has no digit shortcut, so divide honestly. How many win a cookie?

Show the solution

Seven has no digit shortcut, so divide.

2, 051 = 7 \times 293, 4, 613 = 7 \times 659, 8, 267 = 7 \times 1, 181

Those three land exactly, while

5, 000 = 7 \times 714 + 2

and

3, 002 = 7 \times 428 + 6

leave remainders. So

3

tickets win a cookie. Not every divisor comes with a digit test, and for one like 7 you divide and read the remainder.

Practice

A wristband serial reads 41_9 with the tens digit hidden. Every serial is divisible by 11. Use the alternating-sum test backward to find the hidden digit.

Show the solution

Call the hidden tens digit

d

. Reading from the right with the units plus, the alternating sum is

9 - d + 1 - 4 = 6 - d

, and it must be a multiple of 11. Since

d

runs from 0 to 9,

6 - d

runs from 6 down to

- 3

, and the only multiple of 11 there is 0, so

d = 6

. That makes the serial

4169

, and

11 \times 379 = 4169

. The alternating-sum test runs backward the same way the digit-sum test does.

Practice

Since

1001 = 7 \times 11 \times 13

, chop the number into three-digit blocks from the right and alternately subtract and add them. The result shares the same divisibility by 7, 11, and 13.

Tests for Divisibility

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