math.wiki
Number › Divisibility and primes › Listing multiples
Open on the map
Grade 4 · 4.OA.B.4

Listing multiples

Yaretzi jumps rope in rounds of exactly 7 jumps. Which totals can her count ever land on, and could it ever land on 30?

You need
Factors and multiplesGrade 4Skip countingGrade 2
Unlocks
Common multiplesGrade 5Number and shape patterns from a ruleGrade 4
Where you are 10 in Divisibility and primes
Number9
Divisibility and primes10
Factors and multiplesDivisibility rules
Listing multiples
Perfect, abundant, deficient and amicable numbersPrime and composite numbers▾ Show all 10 (5 more)

When you reach for it

Yaretzi jumps rope in rounds. Every round is exactly 7 jumps, and she counts as she goes. After 1 round her count is 7. After 2 rounds it is 14, after 3 rounds 21, and so on. Each round adds 7 more.

0 7 14 21 28 35 42 +7 +7 +7 +7 +7 +7
Figure 1Rounds of 7 jumps. The totals the count can land on are 7, 14, 21, 28, 35, 42 and so on: the multiples of 7. The number 30 sits between two hops and is never landed on.

So which totals can her count be at the end of a round? Only the numbers 7 \times 1, 7 \times 2, 7 \times 3, and so on. These are the multiples of 7. Could her count be 30 at the end of a round? Look at the number line: 7 \times 4 = 28 and 7 \times 5 = 35, and 30 sits between them. No hop lands on 30, so 30 is not a multiple of 7.

Reach for multiples whenever something comes in equal amounts again and again: jumps in a round, beads in a repeat of a pattern, cookies in a box. The total after some number of whole repeats is always a multiple of the amount in one repeat.

The procedure

Part 1: listing the multiples of a number

A multiple of a whole number is that number times a whole number. The multiples of 7 are

7 \times 1,\quad 7 \times 2,\quad 7 \times 3,\quad 7 \times 4,\quad \ldots

which are 7, 14, 21, 28, \ldots Writing them out is Skip counting by 7: counting in equal jumps of 7 instead of jumps of 1.

Why each one is the last one plus the number. Going from 7 \times 4 to 7 \times 5 adds one more group of 7:

7 \times 5 = 7 \times 4 + 7 \times 1 = 28 + 7 = 35

Splitting 5 groups into 4 groups and 1 group like this is the distributive property. It is why the list can be made by adding 7 each time, and it is why the hops on the number line are all the same length.

The list never ends. Any multiple of 7, however big, has another one 7 past it. So there is no largest multiple, and “the multiples of 7” can never all be written down. When a question asks for multiples, it asks for some of them: the first ten, or the ones up to 100, or the ones between two numbers. Ending a written list with “\ldots” says that it keeps going.

The first multiple is the number itself. The list starts with 7 \times 1 = 7. A whole number can also be multiplied by 0, and 7 \times 0 = 0, so 0 is a multiple of 7. In fact 0 is a multiple of every number. It tells us nothing about the number, so lists of multiples usually start at the number itself, as all the lists in this lesson do.

Factor, divisible by, multiple of. In Factors and multiples one fact was said three ways. The fact 7 \times 4 = 28 says 7 is a factor of 28, 28 is divisible by 7, and 28 is a multiple of 7. Listing factors starts from a number and finds what goes into it. Listing multiples goes the other way: it starts from a number and finds what it goes into. A number has only a few factors, but it has endlessly many multiples.

Part 2: deciding whether a number is a multiple

Divide and look at the remainder. A number is a multiple of 7 exactly when 7 goes into it with remainder 0. That is the definition said the other way round: 28 \div 7 = 4 \text{ R } 0 says 28 = 7 \times 4. For 30,

30 \div 7 = 4 \text{ R } 2

and the remainder 2 is the two jumps past 28 that a whole round cannot reach. So 30 is not a multiple of 7.

Or find the two neighbors. Another way to see the same thing is to find the multiples just below and just above the number. If the number is one of them, it is a multiple. If it falls strictly between them, it is not, because the hops jump straight from one to the next.

Multiples on a hundred chart. A hundred chart puts 1 to 100 in rows of ten. Shade the multiples of one number and a pattern appears.

  • The multiples of 10 fill the last column, since each row ends 10 further on.
  • The multiples of 2 fill every other column: they are the even numbers. The last digit alone decides this, as Divisibility rules shows.
  • The multiples of 9 run on a slant, as in the figure below. Adding 9 is the same as adding 10 and taking away 1. Adding 10 moves one row down, and taking away 1 moves one column left, so from 9 to 81 each multiple of 9 sits one row down and one column left of the one before. From 81 there is no column further left: adding 10 gives 91 at the start of the last row, and taking away 1 steps back to 90 at the end of the row above. Then 99 is one row down and one column left of 90, a second short slant.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Figure 2The multiples of 9 on a hundred chart. From 9 to 81 each one sits one row down and one column left of the one before. Then 90 and 99 make a second short slant.

Something else shows up for 9. Add the digits of each multiple: 1 + 8 = 9, 2 + 7 = 9, 6 + 3 = 9. Every multiple of 9 from 9 to 90 has digit sum 9, and 99 has digit sum 9 + 9 = 18. Here this is only noticed, not proved. Why it happens, and how it becomes a test, is in Tests for three and nine.

Stories about things that repeat. A pattern of beads that repeats every 5 beads ends a repeat at bead 5, bead 10, bead 15: at the multiples of 5. A rope counter that beeps every 12 jumps beeps at the multiples of 12. To find what happens at some far-off bead or jump, find the nearest multiple below it, where a repeat has just ended, and count on from there.

Worked examples

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

Example 1

Ten rounds of 6

Sameer jumps rope in rounds of 6 jumps. Write his count at the end of each of the first 10 rounds.

Solution

The count after some number of rounds is 6 times that number, so the counts are the first ten multiples of 6. Start at 6 \times 1 = 6 and add 6 each time, since each round adds one more group of 6:

6,\ 12,\ 18,\ 24,\ 30,\ 36,\ 42,\ 48,\ 54,\ 60

Check the last one by multiplying instead of adding: 6 \times 10 = 60. It matches, so no step of adding went wrong.

Example 2

Every fourth bead

Brynja strings a necklace of 24 beads. Every 4th bead is gold, and the rest are white. Which beads are gold, and how many gold beads are there?

Solution

Counting along the string, bead 4 is gold, then 4 beads later bead 8, then bead 12. The gold beads are at the multiples of 4, and the list stops at 24 because the necklace has 24 beads:

4,\ 8,\ 12,\ 16,\ 20,\ 24
4 8 12 16 20 24
Figure 3Brynja’s necklace of 24 beads with every 4th bead gold. The gold beads sit at the multiples of 4.

That is 6 gold beads. Check: 24 = 4 \times 6, so 24 is the 6th multiple of 4, and the last bead is gold.

Example 3

Multiples of 12 between 50 and 100

List every multiple of 12 that is more than 50 and less than 100.

Solution

First find where to start. Divide: 50 \div 12 = 4 \text{ R } 2, so 12 \times 4 = 48 is just below 50, and the first multiple past 50 is 12 \times 5 = 60. Now add 12 each time:

60,\ 72,\ 84,\ 96

The next one, 96 + 12 = 108, is past 100, so stop. There are 4 multiples of 12 between 50 and 100.

Example 4

Did Sameer finish his rounds?

Sameer only stops at the end of a round of 6 jumps. One day his count is 84. Another day he says it is 92. Could each count be right?

Solution

A count at the end of whole rounds must be a multiple of 6. Test each number by dividing.

84 \div 6 = 14 \text{ R } 0

The remainder is 0, so 84 = 6 \times 14. That count is possible: 14 rounds.

92 \div 6 = 15 \text{ R } 2

The remainder is 2, so 92 is not a multiple of 6. The neighbors are 6 \times 15 = 90 and 6 \times 16 = 96, and 92 falls between them. The count 92 cannot be right; he stopped 2 jumps into a round, or he miscounted.

Example 5

The nines on a hundred chart

Shade the multiples of 9 on a hundred chart, and describe two patterns in them.

Solution

Count by 9 from 9, stopping before passing 100:

9,\ 18,\ 27,\ 36,\ 45,\ 54,\ 63,\ 72,\ 81,\ 90,\ 99

The next one would be 108, past the chart. That is 11 multiples, the picture in Part 2.

First pattern: from 9 to 81 they make a slanting line, each one a row down and a column left of the last, because adding 9 is adding 10 and taking away 1. So the tens digit goes up by 1 and the ones digit goes down by 1 each time, all the way to 90: 18, 27, 36, and so on. On the chart, 90 sits at the right end of its row, since 81 is already in the first column, and 90 and 99 make a second short slant.

Second pattern: the digit sums are all 9 up to 90 (for example 7 + 2 = 9), and the digit sum of 99 is 18. The first pattern explains the second: one digit goes up by 1 while the other goes down by 1, so the sum stays the same.

Example 6

The 38th bead

Brynja’s next necklace repeats the same 5 beads over and over: red, red, white, white, green. What color is the 38th bead?

Solution

One repeat is 5 beads, so a repeat ends with a green bead at bead 5, bead 10, bead 15, at every multiple of 5. Find the multiple of 5 just below 38:

38 \div 5 = 7 \text{ R } 3

So 7 whole repeats end at bead 5 \times 7 = 35, which is green. Then count on 3 more beads from the start of the pattern: bead 36 is red, bead 37 is red, bead 38 is white. The 38th bead is white.

The remainder 3 said the same thing at once: bead 38 is the 3rd bead of a repeat, and the 3rd bead of the pattern is white.

Example 7

A day of rounds

Yaretzi does rounds of 8 jumps. In the morning she does 5 rounds. In the afternoon she does some more whole rounds. She says her total for the day is 100. Can that be right? If not, what are the two possible totals closest to 100?

Solution

Her morning total is 8 \times 5 = 40. The afternoon is some whole number of rounds of 8. If it is 3 rounds, say, the day’s total is

8 \times 5 + 8 \times 3 = 8 \times 8 = 64

by the distributive property: 5 groups of 8 and 3 more groups of 8 make 8 groups of 8. The same thing happens for any number of afternoon rounds, so the day’s total is always 8 times the whole number of rounds: a multiple of 8. In words, every jump of the day belongs to some whole round of 8.

Now test 100: 100 \div 8 = 12 \text{ R } 4. It is not a multiple of 8, so her total cannot be 100. The neighbors are 8 \times 12 = 96 and 8 \times 13 = 104. A total of 96 means 12 - 5 = 7 afternoon rounds, and 104 means 13 - 5 = 8 afternoon rounds. Both are possible, and 100 is not.

Boundaries

  • Whole numbers only. A multiple of 4 is 4 times a whole number. It is true that \displaystyle 4 \times 3\frac{1}{2} = 14, but 14 is not a multiple of 4: 14 \div 4 = 3 \text{ R } 2.
  • The remainder must be exactly 0. Any other remainder, small or large, means “not a multiple”.
  • The list has no end. “List the multiples of 6” always means some of them, up to a limit.
  • Zero. The number 0 is a multiple of every number, since n \times 0 = 0. But the only multiple of 0 is 0 itself, since 0 times any number is 0. The lists in this lesson are for numbers from 1 up, and they start at the number, not at 0.
  • The digit-sum pattern belongs to 9. Digit sums of 9 and 18 were noticed for multiples of 9 only. The multiples of 7, say, have no such pattern: 14 has digit sum 5 and 21 has digit sum 3.

Where this goes next

Lists of multiples help sort numbers in Prime and composite numbers: every multiple of 3 after 3 itself, such as 12 or 21, has 3 as a factor besides 1 and itself, so it has more than two factors. Crossing out such multiples in order is The sieve of Eratosthenes. A list of multiples is also a pattern made from a rule, “start at n and add n”, and Number and shape patterns from a rule studies such patterns for the features the rule does not say. When two numbers each have a list of multiples, the numbers on both lists matter, and finding the smallest one is Least common multiple by listing.

Exercises

Exercises 1 to 9 go with Part 1, and 10 to 21 with Part 2. The reasoning group draws on both.

Part 1: listing the multiples of a number

  1. List the first ten multiples of 3.
  2. List the first eight multiples of 11.
  3. List the first six multiples of 14.
  4. List the multiples of 25 up to 200.
  5. Brynja’s necklace has 30 beads, and every 6th bead is a star bead. Which beads are star beads, and how many are there?
  6. Yaretzi does rounds of 15 jumps. Write her count at the end of each of the first 6 rounds.
  7. List every multiple of 3 that is more than 40 and less than 55.
  8. What is the 10th multiple of 16? What is the 5th?
  9. Sameer says, “The smallest multiple of 5 in a list is 5, and the largest is 50.” Which part is right, and which is wrong?

Part 2: deciding, the hundred chart and stories

  1. Is 72 a multiple of 4? Show the division.
  2. Is 58 a multiple of 6? Show the division.
  3. Is 65 a multiple of 8? Name the two multiples of 8 it falls between, or say which it is.
  4. Which of 34, 45, 56, 63 and 74 are multiples of 7?
  5. Which of 26, 39, 52, 64 and 78 are multiples of 13?
  6. On a hundred chart, where do the multiples of 5 sit, and how many are there from 1 to 100?
  7. How many multiples of 4 are on a hundred chart?
  8. How many multiples of 3 are on a hundred chart, and what is the largest?
  9. Sameer’s rope counter beeps on every 12th jump. He does 100 jumps. How many times does it beep, and on which jump is the last beep?
  10. A bead pattern repeats every 4 beads: blue, blue, yellow, white. What color is the 30th bead?
  11. Brynja has 50 beads and makes bracelets of exactly 8 beads each. Can she use every bead? How many more beads would she need to make one more bracelet with nothing left over?
  12. Yaretzi does whole rounds of 7 jumps. Her count is more than 60 and less than 80. What could it be?

Reasoning

  1. Explain why there is no largest multiple of 6.
  2. Is 0 a multiple of 7? Is 7 a multiple of 0? Explain both.
  3. Brynja says, “Every multiple of 9 has digit sum 9, so 99 is not a multiple of 9.” Show that she is wrong, and say what is true about the digit sums on a hundred chart.
  4. Explain why every multiple of 10 is also a multiple of 5.
Answers
  1. 3, 6, 9, 12, 15, 18, 21, 24, 27, 30. Start at 3 \times 1 and add 3 each time; 3 \times 10 = 30.
  2. 11, 22, 33, 44, 55, 66, 77, 88. The last is 11 \times 8 = 88.
  3. 14, 28, 42, 56, 70, 84. The last is 14 \times 6 = 84.
  4. 25, 50, 75, 100, 125, 150, 175, 200. Add 25 each time; 200 = 25 \times 8, so there are 8.
  5. Beads 6, 12, 18, 24, 30, so 5 star beads. They are the multiples of 6 up to 30, and 30 = 6 \times 5.
  6. 15, 30, 45, 60, 75, 90. The last is 15 \times 6 = 90.
  7. 42, 45, 48, 51, 54. 40 \div 3 = 13 \text{ R } 1, so the first is 3 \times 14 = 42; the next after 54 is 57, too big.
  8. 160 and 80. 16 \times 10 = 160 and 16 \times 5 = 80.
  9. The first part is right: a list of multiples of 5 starts at 5 \times 1 = 5. The second is wrong: 50 = 5 \times 10 is only the 10th, and 55, 60 and so on keep going. There is no largest.
  10. Yes. 72 \div 4 = 18 \text{ R } 0, so 72 = 4 \times 18.
  11. No. 58 \div 6 = 9 \text{ R } 4, since 6 \times 9 = 54 and 6 \times 10 = 60.
  12. No. It falls between 64 and 72: 8 \times 8 = 64 and 8 \times 9 = 72, so 65 \div 8 = 8 \text{ R } 1.
  13. 56 and 63. 56 = 7 \times 8 and 63 = 7 \times 9, while 34 \div 7 = 4 \text{ R } 6, 45 \div 7 = 6 \text{ R } 3 and 74 \div 7 = 10 \text{ R } 4.
  14. 26, 39, 52 and 78. They are 13 \times 2, 13 \times 3, 13 \times 4 and 13 \times 6; 64 \div 13 = 4 \text{ R } 12.
  15. In two columns: the fifth column (numbers ending in 5) and the last column (ending in 0). There are 20, since 5 \times 20 = 100.
  16. 25, since 4 \times 25 = 100 is the last one.
  17. 33, and the largest is 99. 3 \times 33 = 99, and 3 \times 34 = 102 is past 100.
  18. 8 beeps, the last on jump 96. 100 \div 12 = 8 \text{ R } 4, and 12 \times 8 = 96.
  19. Blue. 30 \div 4 = 7 \text{ R } 2, so 7 repeats end at bead 28, and bead 30 is the 2nd of a repeat, which is blue.
  20. No. 50 \div 8 = 6 \text{ R } 2: 6 bracelets use 48 beads and 2 are left. With 6 more beads she has 56 = 8 \times 7, enough for 7 bracelets exactly.
  21. 63, 70 or 77. These are 7 \times 9, 7 \times 10 and 7 \times 11; 7 \times 8 = 56 is too small and 7 \times 12 = 84 too big.
  22. Take any multiple of 6: it is some whole number of groups of 6. Add one more group of 6 and the result is still a whole number of groups of 6, so it is also a multiple of 6, and it is bigger. For example 600 = 6 \times 100, and 606 = 6 \times 101 is bigger. So whatever multiple is named, there is a bigger one, and the list never ends.
  23. 0 is a multiple of 7, because 7 \times 0 = 0 and 0 is a whole number. 7 is not a multiple of 0, because 0 times any whole number is 0, never 7. The only multiple of 0 is 0.
  24. 9 \times 11 = 99, so 99 is a multiple of 9 (99 \div 9 = 11 \text{ R } 0). What is true on the chart: the multiples of 9 from 9 to 90 have digit sum 9, and 99 has digit sum 18. The pattern was only noticed, so it cannot overrule a division.
  25. A multiple of 10 is some whole number of tens. Each ten is two fives, since 10 = 5 \times 2, so that many tens is twice as many fives: a whole number of fives. So it is a multiple of 5. For example 7 tens is 14 fives: 70 = 10 \times 7 = 5 \times 14.
kg1128 · updated 2026-10-20 · CC-BY-SA-4.0