Grade 4

Divisibility rules

A sock drive packs socks in pairs, in bundles of five and in boxes of ten. Can you tell from the count alone whether they will pack with none left over, without dividing?

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When you reach for it

Rashmi runs a sock drive. The socks go out to shelters packed three ways: in pairs, in bundles of 5, and in boxes of 10. One week the drive collects 2{,}346 socks. Will they pack into pairs with none left over? Into bundles of 5? Into boxes of 10?

Each question asks whether 2{,}346 is divisible by a number, which means it divides with remainder 0. In Factors and multiples every such question was a division. This lesson shows that for 2, 5 and 10 one glance at the last digit settles it, and that for 4, 8 and 6 a short look at a few digits does. The answer for 2{,}346 comes from its last digit, 6: the socks pack into pairs, but bundles of 5 leave 1 sock over and boxes of 10 leave 6.

Reach for these tests when a number is too big to divide comfortably in your head and you only need to know whether it packs evenly, or how many are left over. They also speed up finding the factors of a number, since several of the small divisions become a glance.

The procedure

Part 1: the last digit decides 2, 5 and 10

Here are the three tests. Each one looks only at the ones digit, the last digit of the number.

A whole number is divisible by 2 exactly when its last digit is 0, 2, 4, 6 or 8. Those numbers are the even numbers, as in Even and odd numbers. A number ending in 1, 3, 5, 7 or 9 is odd, and pairs leave exactly 1 over.
A whole number is divisible by 5 exactly when its last digit is 0 or 5.
A whole number is divisible by 10 exactly when its last digit is 0.

So 2{,}346 is divisible by 2, since it ends in 6. It is not divisible by 5 or by 10.

Why only the last digit matters. Write the number in Expanded form, as the sum of what each digit is worth:

2{,}346 = 2000 + 300 + 40 + 6

The first three parts, 2000 + 300 + 40 = 2340, are a whole number of tens: 234 tens. So every whole number is some tens plus its last digit:

2{,}346 = 234 \text{ tens} + 6

Now look at one ten. Ten socks make 5 pairs, or 2 bundles of 5, or 1 box of 10, with none left over, because 10 = 2 \times 5 and 10 = 1 \times 10. If one ten packs evenly, then 234 tens pack evenly too: they are just 234 copies of the same packing. So the tens part never leaves a sock over, whatever its digits are. The only socks that can be left over are the ones in the last digit.

Figure 1Every ten splits into 5 pairs, so only the ones can leave a sock over. 27 ends in 7, and 7 is 3 pairs and 1 over.

The picture shows 27 socks, 2 tens and 7 ones. Each ten is 5 pairs. The 7 ones make 3 pairs and 1 over. That 1 is the only sock over in all 27, so 27 is odd.

The leftover is the leftover of the last digit. The same reason gives more than yes or no. Whatever the last digit leaves over is exactly what the whole number leaves over. For bundles of 5, a last digit of 6 is 5 + 1, so 2{,}346 socks leave 1 over. For boxes of 10, the last digit itself is the leftover: 2{,}346 is 234 boxes and 6 socks over.

These three tests work because 2, 5 and 10 each divide 10. That is what lets a whole ten be set aside. Most other numbers do not divide 10, so the last digit cannot settle them. Part 2 shows what to do instead.

Part 2: the last two or three digits for 4 and 8, and two tests for 6

The test for 4. Ten is not a multiple of 4: ten socks in fours make 2 fours and 2 over. So the last digit cannot decide 4. The numbers 12 and 22 both end in 2, and 12 = 4 \times 3 is divisible by 4, but 22 = 4 \times 5 + 2 is not.

One hundred, though, is a multiple of 4:

100 = 4 \times 25

Figure 2One hundred splits into 4 blocks of 25, so 100 = 4 \times 25. Every hundred packs into fours, and only the last two digits are left to test.

Every hundred packs into fours with none left over. So do 1000, 10{,}000 and every bigger place, since each is a whole number of hundreds. Split any number into its hundreds and its last two digits. For 5{,}316:

5{,}316 = 53 \text{ hundreds} + 16

The hundreds part always packs into fours. Only the number made by the last two digits, here 16, can leave anything over.

A whole number is divisible by 4 exactly when the number made by its last two digits is divisible by 4. Since 16 = 4 \times 4, the number 5{,}316 is divisible by 4. As in Part 1, the leftover of the last two digits is the leftover of the whole number.

The test for 8. One hundred is not a multiple of 8: 100 = 8 \times 12 + 4. So two digits are not enough. One thousand is a multiple of 8:

1000 = 8 \times 125

Every thousand, and every bigger place, packs into eights.

A whole number is divisible by 8 exactly when the number made by its last three digits is divisible by 8.

The test for 6. No place packs into sixes: 10 = 6 + 4, 100 = 6 \times 16 + 4, and so on, each leaving 4 over. So no group of last digits settles 6. Instead, use 6 = 2 \times 3 and test twice.

A whole number is divisible by 6 exactly when it is divisible by both 2 and 3. The test for 2 is the last digit, from Part 1. The test for 3 adds up all the digits, as in the test for 3.

Why both together are enough: say a number passes the test for 3, so it is 3 times some whole number. If that whole number were odd, 3 times it would be odd too, since three odd numbers add to an odd number. The number is even, so that whole number must be even, which means it is 2 times something. Then the number is 3 \times 2 \times something, which is 6 \times something. This works because 2 and 3 share no factor except 1, so being divisible by 2 adds something that being divisible by 3 did not already give.

And it goes the other way: a multiple of 6 is 2 \times 3 \times something, so it is divisible by 2 and by 3. If either test fails, the number is not divisible by 6.

Which tests look only at the last digits. The tests for 2, 5 and 10 use the last digit; 4 uses the last two digits; 8 uses the last three. The test for 6 is different: its 3 part uses every digit, and its 2 part uses the last one.

Worked examples

Examples 1, 2 and 3 go with Part 1, and Examples 4 to 7 with Part 2.

Example 1

Pairs, fives and tens

Rashmi’s drive collects 2{,}735 socks. Will they pack into pairs, into bundles of 5, and into boxes of 10 with none left over? For each one that leaves socks over, say how many.

Solution

Look at the last digit, 5.

Pairs: 5 is odd, so 2{,}735 is odd and pairs leave 1 sock over. Check: 2 \times 1367 = 2734, and 2735 - 2734 = 1.

Bundles of 5: the last digit is 5, so 2{,}735 is divisible by 5. There are 2735 \div 5 = 547 bundles, none over.

Boxes of 10: the last digit is not 0, so boxes leave socks over. The tens part is 273 tens, so there are 273 full boxes and the last digit, 5, is the number of socks over.

Example 2

Seeing why with expanded form

Kusi counts 4{,}068 socks for the drive. Use expanded form to show that they pack into pairs, and find how many pairs there are without long division.

Solution

In expanded form, 4{,}068 = 4000 + 60 + 8, since the hundreds digit is 0. The first two parts make 4060, which is 406 tens. So

4{,}068 = 406 \text{ tens} + 8

Each ten is 5 pairs, so 406 tens are 406 \times 5 = 2030 pairs. The last 8 socks make 4 pairs. Nothing is left over, so the socks pack into pairs, as the test says for a last digit of 8. The number of pairs is

2030 + 4 = 2034

Check: 2 \times 2034 = 4068.

Example 3

How many more to fill the last bundle

Vebjorn has 1{,}483 socks. How many are left over in bundles of 5, and how many more socks would fill one more bundle? Answer the same two questions for boxes of 10.

Solution

Split off the tens: 1{,}483 = 148 \text{ tens} + 3. The tens part packs evenly both ways, so only the 3 matters.

Bundles of 5: the 3 last socks are less than 5, so 3 are left over. Each ten is 2 bundles, so the tens make 148 \times 2 = 296 bundles. To fill one more bundle he needs 5 - 3 = 2 more socks. Check: 5 \times 296 = 1480, and 1480 + 3 = 1483.

Boxes of 10: the last digit is the leftover, so 3 are over, after 148 full boxes. To fill one more box he needs 10 - 3 = 7 more socks.

Example 4

Towers of four shoe boxes

At the shoe store, Kusi stacks shoe boxes in towers of 4. One delivery has 7{,}324 boxes, and another has 7{,}342. Which delivery stacks with no box left over?

Solution

The test for 4 uses the number made by the last two digits.

For 7{,}324 the last two digits make 24, and 24 = 4 \times 6, so 7{,}324 is divisible by 4. To see why, split off the hundreds:

7{,}324 = 73 \text{ hundreds} + 24

Each hundred is 25 fours, so 73 hundreds are 73 \times 25 = 1825 fours, and 24 is 6 more. That makes 1825 + 6 = 1831 towers. Check: 4 \times 1831 = 7324.

For 7{,}342 the last two digits make 42, and 42 = 4 \times 10 + 2. So 7{,}342 is not divisible by 4, and 2 boxes are left over. The two numbers use the same digits, but the order of the last two changes the answer. Notice too that both numbers end in an even digit, so the last digit alone could not tell them apart.

Example 5

Cartons of eight

Sneakers ship to the shoe store in cartons of 8 pairs. Vebjorn has 6{,}136 pairs to pack, and Rashmi has 6{,}236. Whose pairs fill their cartons exactly?

Solution

The test for 8 uses the number made by the last three digits.

For 6{,}136 that number is 136, and 136 = 8 \times 17. So 6{,}136 is divisible by 8. Split off the thousands to count the cartons: 6{,}136 = 6 \text{ thousands} + 136, and each thousand is 125 cartons, so there are 6 \times 125 + 17 = 750 + 17 = 767 cartons. Check: 8 \times 767 = 6136.

For 6{,}236 that number is 236, and 236 = 8 \times 29 + 4. So 6{,}236 is not divisible by 8, and 4 pairs are left over.

Both numbers end in 36, and 36 = 4 \times 9, so both are divisible by 4. The last two digits settle 4 but not 8, because a hundred is 12 eights and 4 over, so the hundreds digit changes the leftover.

Example 6

Packs of six

The drive also sends socks in packs of 6. Which of these counts pack into sixes with none left over: 2{,}814, 3{,}254 and 1{,}827?

Solution

A number is divisible by 6 exactly when it passes both the test for 2 and the test for 3.

2{,}814: the last digit 4 is even, so it passes the test for 2. Its digit sum is 2 + 8 + 1 + 4 = 15, and 15 = 3 \times 5, so it passes the test for 3. Both pass, so 2{,}814 is divisible by 6. To see the reason in this case, 2814 = 3 \times 938, and 938 is even, 938 = 2 \times 469, so 2814 = 3 \times 2 \times 469 = 6 \times 469. That is 469 packs.

3{,}254: it is even, but its digit sum is 3 + 2 + 5 + 4 = 14, which is not a multiple of 3. It fails the test for 3, so it is not divisible by 6. Check: 6 \times 542 = 3252, so 2 socks are over.

1{,}827: its digit sum is 1 + 8 + 2 + 7 = 18, so it passes the test for 3. But it ends in 7, so it is odd and fails the test for 2. It is not divisible by 6. Check: 6 \times 304 = 1824, so 3 socks are over.

Only 2{,}814 packs into sixes. Passing one of the two tests is never enough.

Example 7

Every pack size from 2 to 10

Rashmi has 2{,}520 socks and wants to pack all of them in equal packs. Which pack sizes from 2 to 10 use every sock? Use the tests wherever one applies.

Solution

Go through the sizes, using the digits of 2{,}520.

The last digit is 0, so 2, 5 and 10 all work.
The last two digits make 20 = 4 \times 5, so 4 works.
The last three digits make 520 = 8 \times 65, so 8 works.
The digit sum is 2 + 5 + 2 + 0 = 9, so 3 works, and 9 works too, by the test in Tests for three and nine.
It passes the tests for 2 and 3, so 6 works.
This lesson has no test for 7, so divide: 7 \times 360 = 2520, so 7 works as well.

Every size from 2 to 10 uses all the socks. The tests settled eight of the nine sizes with no division at all. The quotients, which count the packs, are

\begin{aligned} &2520 \div 2 = 1260, \quad 2520 \div 3 = 840, \\ &2520 \div 4 = 630, \quad 2520 \div 5 = 504, \\ &2520 \div 6 = 420, \quad 2520 \div 8 = 315, \\ &2520 \div 9 = 280, \quad 2520 \div 10 = 252 \end{aligned}

So 2, 3, 4, 5, 6, 7, 8, 9 and 10 are all factors of 2{,}520, found mostly by looking at digits.

Boundaries

Whole numbers written in tens. Every test here rests on place value: a ten, a hundred and a thousand are the pieces that pack evenly. The tests apply to whole numbers written with the usual digits 0 to 9.
Only these tests use only the last digits. The last digit settles 2, 5 and 10, and nothing else, because those are the numbers from 2 up that divide 10. The last two digits settle 4 (since 100 = 4 \times 25), and the last three settle 8 (since 1000 = 8 \times 125). For 3, 6 and 9 no group of last digits is enough: 12 and 22 end in the same digit, and 12 and 112 end in the same two digits, but only 12 is divisible by 3 and by 6.
Two tests make a new one only when the numbers share no factor except 1. Divisible by 2 and by 3 means divisible by 6, and divisible by 2 and by 5 means divisible by 10. The same is false for 2 and 4, which share the factor 2: the numbers 4, 12 and 20 are each divisible by 2 and by 4, but none is divisible by 8. Every multiple of 4 is already even, so the test for 2 adds nothing.
Leftovers. For 2, 5, 10, 4 and 8, the leftover of the last digits is the leftover of the whole number. The test for 6 answers only yes or no; to find the leftover, divide.
Zero. The number 0 ends in 0 and is divisible by every one of these numbers, since 0 = 6 \times 0, for example. Zero socks pack with none left over.

Where this goes next

These tests make factor lists faster, and they sort numbers in Prime and composite numbers: a number from 3 up that is even, or ends in 5 and is bigger than 5, has a factor other than 1 and itself, so it is not prime. Reading the last digit is also a quick way to see which numbers appear when you list Listing multiples: multiples of 5 end in 0 or 5, multiples of 10 in 0. The digit-sum tests for 3 and 9 come next in Tests for three and nine.

Later the same reasoning is put in one language. In Modular arithmetic only the leftover counts, and saying “a ten leaves nothing over when packed in fives” becomes a short statement about remainders. Where the divisibility tests come from uses that language to explain every test here, and the tests for 3 and 9, as one idea.

Exercises

Exercises 1 to 11 go with Part 1, and 12 to 22 with Part 2. The reasoning group draws on both.

Part 1: the last digit

Which of 2, 5 and 10 divide 3{,}470?
Which of 2, 5 and 10 divide 8{,}615?
Which of 2, 5 and 10 divide 9{,}052?
Which of 2, 5 and 10 divide 6{,}709?
Kusi packs 4{,}326 socks in bundles of 5. How many socks are left over?
Rashmi packs 1{,}999 socks in pairs. How many pairs does she make, and how many socks are left over?
Vebjorn packs 5{,}284 socks in boxes of 10. How many full boxes are there, and how many more socks would fill one more box?
Write 738 in expanded form, and use it to find how many pairs 738 socks make without long division.
A sock count is 4{,}81\square, with the last digit missing. The socks pack into pairs and into bundles of 5. What is the last digit?
Vebjorn’s sock count is between 70 and 80. It packs into bundles of 5 but not into boxes of 10. What is the count?
Which of 230, 345, 512 and 1{,}005 leave a sock over in pairs but none over in bundles of 5?

Part 2: four, eight and six

Is 3{,}516 divisible by 4?
Is 7{,}154 divisible by 4? If not, what is left over?
Is 9{,}320 divisible by 8?
Is 5{,}428 divisible by 8? Is it divisible by 4?
Is 4{,}578 divisible by 6?
Is 2{,}375 divisible by 6?
Is 5{,}134 divisible by 6?
Kusi stacks 1{,}236 shoe boxes in towers of 4. Are any boxes left over? How many towers are there?
The drive has 3{,}042 socks to send in packs of 6. Do they pack with none left over? How many packs?
Which of 2, 3, 4, 5, 6, 8 and 10 are factors of 2{,}610?
The shoe store’s count of sneaker pairs is 7{,}31\square. Which last digits make the count divisible by 4? Which make it divisible by 8?

Reasoning

Kusi says, “4{,}532 ends in 2, which is even, so 4{,}532 is divisible by 4.” Is 4{,}532 divisible by 4? Is her reason a good one? Explain.
Vebjorn says, “A number divisible by 2 and by 3 is divisible by 6, so a number divisible by 2 and by 4 must be divisible by 8.” Show that he is wrong, and explain why the first rule works and the second does not.
Explain why the test for 8 looks at the last three digits and not just the last two. Use 1{,}100 and 1{,}200, which end in the same two digits.

Answers

All three. It ends in 0: 3470 \div 2 = 1735, 3470 \div 5 = 694, 3470 \div 10 = 347.
Only 5. It ends in 5, so it is odd and does not end in 0. 8615 \div 5 = 1723.
Only 2. It ends in 2, which is even, but not 0 or 5. 9052 \div 2 = 4526.
None of them. It ends in 9: odd, and not 0 or 5.
1 sock. The last digit 6 is 5 + 1. Check: 5 \times 865 = 4325, one less than 4326.
999 pairs and 1 sock over. The last digit 9 is odd, and 2 \times 999 = 1998.
528 full boxes and 4 socks over, so 6 more socks fill another box. 5{,}284 is 528 tens and 4, and 10 - 4 = 6.
738 = 700 + 30 + 8, which is 73 tens and 8. The tens make 73 \times 5 = 365 pairs and the 8 makes 4 pairs, so 369 pairs. Check: 2 \times 369 = 738.
0. Bundles of 5 need a last digit of 0 or 5, and pairs need it even, so the count is 4{,}810.
75. The only number between 70 and 80 that ends in 0 or 5 is 75, and it ends in 5, not 0, so it does not fill boxes of 10.
345 and 1{,}005. Both end in 5: odd, and divisible by 5. 230 is even, and 512 does not end in 0 or 5.
Yes. The last two digits make 16 = 4 \times 4. 3516 \div 4 = 879.
No, 2 are left over. The last two digits make 54 = 4 \times 13 + 2. Check: 4 \times 1788 = 7152, and 7154 - 7152 = 2.
Yes. The last three digits make 320 = 8 \times 40. 9320 \div 8 = 1165.
Not by 8: the last three digits make 428 = 8 \times 53 + 4, so 4 are left over. But it is divisible by 4, since the last two digits make 28 = 4 \times 7.
Yes. It ends in 8, even, and its digit sum is 4 + 5 + 7 + 8 = 24, a multiple of 3. 4578 \div 6 = 763.
No. It ends in 5, so it is odd and fails the test for 2.
No. It is even, but its digit sum is 5 + 1 + 3 + 4 = 13, not a multiple of 3. Check: 6 \times 855 = 5130, which leaves 4.
None are left over, and there are 309 towers. The last two digits make 36 = 4 \times 9, and 1236 \div 4 = 309.
Yes, 507 packs. It is even, and its digit sum is 3 + 0 + 4 + 2 = 9, a multiple of 3. 3042 \div 6 = 507.
2, 3, 5, 6 and 10. It ends in 0, so 2, 5 and 10. Its digit sum is 9, so 3, and with 2 that gives 6. The last two digits make 10, not a multiple of 4, so not 4; and a number not divisible by 4 cannot be divisible by 8 (the last three digits make 610 = 8 \times 76 + 2).
For 4: 2 or 6, since the last two digits must be 12 or 16 (the multiples of 4 from 10 to 19). For 8: only 2, since the last three digits must be a multiple of 8 from 310 to 319, and the only one is 312 = 8 \times 39.
It is divisible by 4, because its last two digits make 32 = 4 \times 8. But her reason is not a good one. A last digit cannot settle 4, because a ten is 2 fours and 2 over. The number 4{,}522 also ends in 2, and its last two digits make 22 = 4 \times 5 + 2, so it is not divisible by 4.
12 is divisible by 2 and by 4, but 12 = 8 + 4, so it is not divisible by 8. The first rule works because 2 and 3 share no factor except 1: an even multiple of 3 is 3 times an even number, which is 6 times a whole number. The numbers 2 and 4 share the factor 2, and every multiple of 4 is already even, so passing the test for 2 adds nothing to passing the test for 4.
A hundred is not a multiple of 8: 100 = 8 \times 12 + 4. So the hundreds digit can change the leftover, and two digits are not enough. 1{,}100 and 1{,}200 both end in 00, but 1100 = 8 \times 137 + 4 is not divisible by 8 and 1200 = 8 \times 150 is. A thousand is 8 \times 125, so every place from the thousands up packs into eights, and only the last three digits are left to test.