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Grade 4 · 4.OA.B.4

Factors and multiples

One hundred chairs are set out for the school photo, with every row the same length. In how many different ways can the rows go?

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When you reach for it

Oddny is setting out chairs for the school photo. There are 100 chairs, and she wants every row to have the same number of chairs, with no chair left over. Can she make 4 rows? Can she make 3 rows? How many different ways are there?

Four rows works: 4 rows of 25 chairs use all 100, because 4 \times 25 = 100. Three rows does not: 3 \times 33 = 99, so 3 rows of 33 leave 1 chair over, and 3 rows of 34 would need 102. The numbers of rows that work are exactly the numbers that divide 100 with nothing left over. Those numbers are the factors of 100.

1 row of 100 2 rows of 50 4 rows of 25 5 rows of 20 10 rows of 10
Figure 1100 chairs in equal rows: 1 \times 100, 2 \times 50, 4 \times 25, 5 \times 20 and 10 \times 10. Turning an array on its side gives 20 \times 5, 25 \times 4, 50 \times 2 and 100 \times 1.

Reach for factors whenever a number of things must be split into equal groups, rows or teams with nothing left over, and you want to know every way it can be done. Chairs in rows, singers on choir risers, a class split into equal teams: each asks for the factors of a number.

The procedure

Part 1: a factor leaves remainder 0

In Remainders and what they mean a division gave a quotient and a remainder. Most of the time something is left over. Sometimes the remainder is 0, and that is the case this lesson is about.

A factor of a whole number is a whole number that divides it with remainder 0. Since 100 \div 4 = 25 \text{ R } 0, the number 4 is a factor of 100. Since 100 \div 3 = 33 \text{ R } 1, the number 3 is not.

A remainder of 0 means the groups fit exactly, so the division can be written as a multiplication with no leftover:

100 = 4 \times 25

Whenever two whole numbers multiply to make a number, both of them are factors of it. The fact 4 \times 25 = 100 says 4 is a factor of 100, and it says 25 is a factor of 100 too. In chairs, 4 rows of 25 turned on their side are 25 rows of 4: the same 100 chairs. This is the commutative property of multiplication, 4 \times 25 = 25 \times 4, and it is why factors always come two at a time.

One fact, three ways to say it. People say the fact 100 = 4 \times 25 in three ways, and all three mean exactly the same thing:

SentenceWhat it stresses
4 is a factor of 100.4 goes into 100 exactly.
100 is divisible by 4.dividing 100 by 4 leaves remainder 0
100 is a multiple of 4.counting by 4s lands on 100: 4, 8, 12, \ldots, 96, 100

If one of the three is true, all three are true. If one is false, all three are false: 3 is not a factor of 100, 100 is not divisible by 3, and 100 is not a multiple of 3, since counting by 3s goes 96, 99, 102 and jumps over 100. Listing many multiples of one number is its own lesson, Listing multiples.

An array shows it. Set chairs in rows. If the chairs fill a rectangle with no gaps and none left over, the number of rows and the number in each row are both factors of the number of chairs. If a chair is left over, the number of rows you tried is not a factor. That is why the picture of a factor is an array, as in Arrays.

Two factors every number has. Every whole number from 1 up has 1 as a factor, because n = 1 \times n: one row with all the chairs in it. And every number is a factor of itself, for the same reason. So 1 and the number itself are always on the list. A number whose only factors are 1 and itself, such as 13, is called prime; that idea belongs to Prime and composite numbers.

Part 2: finding every factor pair

A factor pair of a number is two factors that multiply to make it. For 100, 4 and 25 are a factor pair. To find all the factors, find all the pairs. Test 1, then 2, then 3, and so on. Each test is a division. If the remainder is 0, write down the pair: the number you tested and the quotient. If not, move on.

TestDivisionPair
1100 \div 1 = 1001 \times 100
2100 \div 2 = 502 \times 50
3100 \div 3 = 33 \text{ R } 1none
4100 \div 4 = 254 \times 25
5100 \div 5 = 205 \times 20
6100 \div 6 = 16 \text{ R } 4none
7100 \div 7 = 14 \text{ R } 2none
8100 \div 8 = 12 \text{ R } 4none
9100 \div 9 = 11 \text{ R } 1none
10100 \div 10 = 1010 \times 10

At 10 the two numbers in the pair are the same. The pairs have met, and we stop. The five factor pairs of 100 are

\begin{aligned} &1 \times 100, \quad 2 \times 50, \quad 4 \times 25, \\ &5 \times 20, \quad 10 \times 10 \end{aligned}

To read off the factors in order, take the small numbers going down the left side, then the partners going back up the right side:

1,\ 2,\ 4,\ 5,\ 10,\ 20,\ 25,\ 50,\ 100

That is 9 factors, and each one is a number of rows Oddny can use: 1 row of 100, 2 rows of 50, and so on up to 100 rows of 1.

When the pairs cross instead of meet. Most numbers have no pair like 10 \times 10. For those, keep testing until the number you are about to test is already on the list as a partner, or is bigger than the partner it would need. Then the pairs have crossed, and you stop. Example 3 stops at a partner already on the list, and Example 4 stops where the pairs cross. A quick way to tell that you have crossed: the number you are about to test, times itself, is more than the number you are factoring.

Each division only has to be tried once, and only up to the stopping point, so even a number up to 100 needs at most 10 tests. The shortcuts in Divisibility rules make some of those tests one glance at the last digit.

Part 3: why the list stops there, and square numbers

Why stopping is safe. For 100 we stopped at 10 and never tested 11, 12, 13 and the rest. Could a factor have been missed up there? Think about any factor pair of 100. Both numbers in it cannot be bigger than 10, because two numbers bigger than 10 multiply to more than 10 \times 10 = 100; even 11 \times 11 is already 121. So every pair has a number that is 10 or less. All of those were tested, so every pair was already found, and with it every big factor such as 20, 25 and 50. The same reason works for any number: past the meeting or crossing point, every factor would need a partner that has already been tested.

Why a square number has an odd count of factors. Factors come in pairs, two different numbers each time, so a count of factors is usually even. The one exception is a pair whose two numbers are the same, like 10 \times 10. That pair adds only one number to the list. A number made as a whole number times itself is a square number: 100 = 10 \times 10 is one, and its array is a square, 10 rows of 10 chairs. So, for every whole number from 1 up:

  • a square number has an odd count of factors (100 has 4 pairs of two plus the single 10, so 4 \times 2 + 1 = 9 factors, and 1 = 1 \times 1 has the single factor 1);
  • every other number has an even count, because all its pairs are two different numbers.

Factor questions in stories. A story usually adds a condition: at least 3 rows, no more than 10 chairs in a row, rows of more than 5 singers. The method stays the same. List the factor pairs first, read each one as rows and chairs per row (both ways round), then keep only the pairs that fit every condition. Examples 6 and 7 do this.

Worked examples

Examples 1 and 2 go with Part 1, Examples 3 and 4 with Part 2, and Examples 5, 6 and 7 with Part 3.

Example 1

Rows of 7 and rows of 8

The choir has 42 singers. Wayra wants them on the risers in full rows of 7. Can she do it? Could she use full rows of 8 instead? Say each answer with “factor”, “divisible by” and “multiple of”.

Solution

Rows of 7: the fact 7 \times 6 = 42 means 6 rows use every singer, so

42 \div 7 = 6 \text{ R } 0

The remainder is 0, so yes. In the three ways: 7 is a factor of 42; 42 is divisible by 7; 42 is a multiple of 7. The same fact shows 6 is a factor of 42 too.

Rows of 8: the facts nearby are 8 \times 5 = 40 and 8 \times 6 = 48. Five full rows use 40 singers and leave 2:

42 \div 8 = 5 \text{ R } 2

The remainder is not 0, so no. 8 is not a factor of 42; 42 is not divisible by 8; 42 is not a multiple of 8.

Example 2

Which numbers of rows work for 15 chairs

Omkar has 15 chairs for a small photo. Which numbers of equal rows use every chair?

Solution

Try each number of rows from 1 to 15; more than 15 rows would leave some row empty.

  • 1 row: 15 \div 1 = 15. Works.
  • 2 rows: 15 \div 2 = 7 \text{ R } 1. One chair is left over, as the picture shows.
  • 3 rows: 15 \div 3 = 5. Works: 3 rows of 5.
  • 4 rows: 15 \div 4 = 3 \text{ R } 3. No.
  • 5 rows: 15 \div 5 = 3. Works: 5 rows of 3.
  • 6 rows and 7 rows: 15 \div 6 = 2 \text{ R } 3 and 15 \div 7 = 2 \text{ R } 1. No.
  • 8 to 14 rows: each row could hold only 1 chair, with some left over. No.
  • 15 rows: 15 \div 15 = 1. Works.
2 rows, 1 left over 3 rows of 5
Figure 215 chairs in 2 rows leaves 1 chair over, so 2 is not a factor of 15. In 3 rows they fit exactly, so 3 is a factor of 15.

The rows that work are 1, 3, 5 and 15, so the factors of 15 are 1, 3, 5 and 15. Notice they came in pairs: 3 rows of 5 and 5 rows of 3 are the same array turned. That is the shortcut Part 2 uses.

Example 3

All the factors of 90

Find every factor of 90.

Solution

Test 1, 2, 3, \ldots and record each pair.

TestDivisionPair
190 \div 1 = 901 \times 90
290 \div 2 = 452 \times 45
390 \div 3 = 303 \times 30
490 \div 4 = 22 \text{ R } 2none
590 \div 5 = 185 \times 18
690 \div 6 = 156 \times 15
790 \div 7 = 12 \text{ R } 6none
890 \div 8 = 11 \text{ R } 2none
990 \div 9 = 109 \times 10

The next number to test is 10, and 10 is already on the list as the partner of 9. The pairs have met side by side, so stop. Testing 10 would only give 10 \times 9, the same pair turned round. The factors, small ones down and partners back up, are

1,\ 2,\ 3,\ 5,\ 6,\ 9,\ 10,\ 15,\ 18,\ 30,\ 45,\ 90

That is 6 pairs, so 12 factors. Check: 90 is not a square number, and 12 is even, as Part 3 says it must be.

Example 4

An odd number: the factors of 75

Find every factor of 75.

Solution

Start with 1 \times 75. The number 75 is odd, so no even number is a factor: an even number of rows of any length always makes an even number of chairs. That rules out 2, 4, 6 and 8 without dividing. The odd numbers:

  • 3: 75 \div 3 = 25, so 3 \times 25.
  • 5: 75 \div 5 = 15, so 5 \times 15.
  • 7: 7 \times 10 = 70 and 7 \times 11 = 77, so 75 \div 7 = 10 \text{ R } 5. None.
  • 9: 9 \times 8 = 72 and 9 \times 9 = 81, so 75 \div 9 = 8 \text{ R } 3. None.

The next number to test is 10, and 10 \times 10 = 100 is more than 75. Any partner of 10 would have to be less than 10, and every number below 10 has been tested. The pairs have crossed, so stop. The factors are

1,\ 3,\ 5,\ 15,\ 25,\ 75

That is 3 pairs, 6 factors.

Example 5

Counting the factors of 64

How many factors does 64 have? Explain why the count is odd.

Solution

Test in order. 1 \times 64 and 2 \times 32. Then 64 \div 3 = 21 \text{ R } 1, none. Then 4 \times 16. Then 64 \div 5 = 12 \text{ R } 4, 64 \div 6 = 10 \text{ R } 4 and 64 \div 7 = 9 \text{ R } 1, none. Then 8 \times 8 = 64, and the pairs meet. The factors are

1,\ 2,\ 4,\ 8,\ 16,\ 32,\ 64

so 64 has 7 factors. The pairs 1 \times 64, 2 \times 32 and 4 \times 16 each give two different factors, which makes 6. The last pair, 8 \times 8, gives only one new factor, the 8. So the count is 3 \times 2 + 1 = 7, an odd number, because 64 is a square number: 8 rows of 8 chairs make a square.

Example 6

Rows for the class photo

Oddny has 32 chairs for a class photo. She wants at least 3 rows, with the same number of chairs in every row and at least 5 chairs in each row. How should she set them out?

Solution

First find the factor pairs of 32. They are 1 \times 32, 2 \times 16 and 4 \times 8. The tests 32 \div 3 = 10 \text{ R } 2, 32 \div 5 = 6 \text{ R } 2, 32 \div 6 = 5 \text{ R } 2 and 32 \div 7 = 4 \text{ R } 4 all leave remainders, and 8 is already a partner, so those are all the pairs.

Each pair can be read both ways round, as rows and chairs per row:

RowsChairs per rowAt least 3 rows?At least 5 per row?
132noyes
216noyes
48yesyes
84yesno
162yesno
321yesno

Only one way passes both conditions: 4 rows of 8 chairs. Check: 4 \times 8 = 32.

Example 7

Which numbers up to 30 have an odd count of factors

Without listing every factor of every number, find all the whole numbers from 1 to 30 that have an odd number of factors.

Solution

Factors come in pairs of two different numbers, except for a pair whose two numbers are the same. So a number has an odd count of factors exactly when it is a square number, a whole number times itself. List the squares up to 30:

\begin{aligned} &1 \times 1 = 1, \quad 2 \times 2 = 4, \quad 3 \times 3 = 9, \\ &4 \times 4 = 16, \quad 5 \times 5 = 25 \end{aligned}

The next one, 6 \times 6 = 36, is past 30. So the numbers are 1, 4, 9, 16 and 25.

Check two of them. The factors of 16 are 1, 2, 4, 8, 16: five, odd. The number 1 has the single factor 1: one, odd. And a number that is not a square, such as 20, has 1 \times 20, 2 \times 10, 4 \times 5: six factors, even.

Boundaries

  • Whole numbers only. A factor is a whole number, and so is its partner. It is true that \displaystyle 100 = 8 \times 12\frac{1}{2}, but 8 is not a factor of 100: 100 \div 8 = 12 \text{ R } 4.
  • The remainder must be exactly 0. A remainder of 1 is as much a “no” as a remainder of 7.
  • Factors of a number from 1 up lie between 1 and the number. No factor is bigger than the number itself, since even one row of the bigger length would need more chairs than there are.
  • The number 1 has exactly one factor, itself, since its only pair is 1 \times 1.
  • Zero is left out. Division by 0 is not defined, so 0 is never a factor. And every whole number from 1 up divides 0 with remainder 0 (for example 0 = 5 \times 0), so 0 would have no end to its list. This lesson factors only whole numbers from 1 up, mostly from 1 to 100.
  • The stopping point needs the tests in order. Stopping where the pairs meet or cross is safe only because every smaller number was tested first.

Where this goes next

The factor list is the starting point for much of number work. Prime and composite numbers sorts the whole numbers by how many factors they have: exactly two, or more than two. The shortcuts in Divisibility rules and, later, Tests for three and nine decide some tests by looking at digits instead of dividing.

Two numbers can share factors. The largest shared one, found by writing both lists, is Greatest common factor by listing, and it is the tool for splitting two amounts into identical groups. Breaking a number into factors again and again until no more splitting is possible is Prime factorization, and the fact that this always ends in the same pieces is Every number has exactly one prime factorization. Adding up a number’s factors leads to perfect and amicable numbers. “Divides with remainder 0” itself becomes a relation between numbers with rules of its own in Divisibility as a relation.

Exercises

Exercises 1 to 8 go with Part 1, 9 to 16 with Part 2, and 17 to 22 with Part 3. The reasoning group draws on all three.

Part 1: a factor leaves remainder 0

  1. Is 6 a factor of 54? Show the division.
  2. Is 7 a factor of 50? Show the division.
  3. Is 91 divisible by 7?
  4. Is 74 a multiple of 4?
  5. Write “5 is a factor of 35” in the two other ways.
  6. Omkar sets out 21 chairs in equal rows with none left over. List every arrangement.
  7. Which of 2, 3, 4, 5 and 6 are factors of 40?
  8. Wayra says, “9 is a factor of 3, because 9 \div 3 = 3.” Fix her sentence, and say whether 9 is a factor of 3.

Part 2: every factor pair

  1. Find all the factors of 34.
  2. Find all the factors of 52.
  3. Find all the factors of 66.
  4. Find all the factors of 80.
  5. Find all the factors of 99.
  6. Find all the factors of 98.
  7. Find all the factors of 39.
  8. Oddny lists the factors of 70 as 1, 2, 5, 7, 10, 35, 70. Is her list complete? If not, find what is missing.

Part 3: square numbers and stories

  1. How many factors does 49 have? Why is the count odd?
  2. How many factors does 81 have?
  3. Omkar has 33 chairs for the choir risers. He wants equal rows, at least 2 rows, and at least 2 chairs in each row. What can he do?
  4. There are 56 singers. The risers have room for at most 8 rows, and every row must have the same number of singers. In how many ways can the singers stand?
  5. Wayra has 44 chairs for the class photo. She wants equal rows of more than 5 chairs but fewer than 15 chairs. How many rows does she make, and how long are they?
  6. Is there any whole number from 51 to 59 with an odd number of factors? Explain.

Reasoning

  1. Omkar knows that 7 \times 9 = 63. Explain why this one fact shows that 7 and 9 are both factors of 63, and say it with “divisible by” and “multiple of”.
  2. Without listing every factor, explain why 25 has an odd number of factors and 26 has an even number.
  3. Oddny says, “A bigger number always has more factors.” Use numbers from this lesson to show she is wrong, and explain.
Answers
  1. Yes. 54 \div 6 = 9 \text{ R } 0, since 6 \times 9 = 54.
  2. No. 50 \div 7 = 7 \text{ R } 1, since 7 \times 7 = 49 and 7 \times 8 = 56.
  3. Yes. 7 \times 13 = 91, so 91 \div 7 = 13 \text{ R } 0.
  4. No. 74 \div 4 = 18 \text{ R } 2, since 4 \times 18 = 72 and 4 \times 19 = 76.
  5. “35 is divisible by 5” and “35 is a multiple of 5”. All three say 35 = 5 \times 7.
  6. 1 row of 21, 3 rows of 7, 7 rows of 3, and 21 rows of 1. The factors of 21 are 1, 3, 7, 21; 2, 4, 5 and 6 rows leave chairs over (21 \div 2 = 10 \text{ R } 1, 21 \div 4 = 5 \text{ R } 1, 21 \div 5 = 4 \text{ R } 1, 21 \div 6 = 3 \text{ R } 3).
  7. 2, 4 and 5. 40 = 2 \times 20 = 4 \times 10 = 5 \times 8, but 40 \div 3 = 13 \text{ R } 1 and 40 \div 6 = 6 \text{ R } 4.
  8. The division 9 \div 3 = 3 shows that 3 is a factor of 9. 9 is not a factor of 3: 3 \div 9 = 0 \text{ R } 3, and no factor of 3 can be bigger than 3.
  9. 1, 2, 17, 34. Pairs 1 \times 34 and 2 \times 17; 3, 4, 5 leave remainders, and 6 \times 6 = 36 is more than 34, so stop.
  10. 1, 2, 4, 13, 26, 52. Pairs 1 \times 52, 2 \times 26, 4 \times 13; 3, 5, 6, 7 leave remainders, and 8 \times 8 = 64 is more than 52.
  11. 1, 2, 3, 6, 11, 22, 33, 66. Pairs 1 \times 66, 2 \times 33, 3 \times 22, 6 \times 11; 4, 5, 7, 8 leave remainders, and 9 \times 9 = 81 is more than 66.
  12. 1, 2, 4, 5, 8, 10, 16, 20, 40, 80. Pairs 1 \times 80, 2 \times 40, 4 \times 20, 5 \times 16, 8 \times 10; 3, 6, 7 leave remainders, and 9 \times 9 = 81 is more than 80, so stop.
  13. 1, 3, 9, 11, 33, 99. Pairs 1 \times 99, 3 \times 33, 9 \times 11; 99 is odd, so no even factor, 5 and 7 leave remainders, and 10 \times 10 = 100 is more than 99.
  14. 1, 2, 7, 14, 49, 98. Pairs 1 \times 98, 2 \times 49, 7 \times 14; 3, 4, 5, 6, 8, 9 leave remainders, and 10 \times 10 = 100 is more than 98.
  15. 1, 3, 13, 39. Pairs 1 \times 39 and 3 \times 13; 39 is odd, 5 leaves a remainder, and 7 \times 7 = 49 is more than 39.
  16. Not complete: 14 is missing. 70 \div 5 = 14, so 5 \times 14 is a pair, but only the 5 was written. The full list is 1, 2, 5, 7, 10, 14, 35, 70.
  17. 3 factors: 1, 7, 49. The pairs are 1 \times 49 and 7 \times 7, and 7 \times 7 adds only one number, because 49 is a square number.
  18. 5 factors: 1, 3, 9, 27, 81, from 1 \times 81, 3 \times 27 and 9 \times 9.
  19. 3 rows of 11 chairs, or 11 rows of 3 chairs. The factor pairs of 33 are 1 \times 33 and 3 \times 11, and 1 row or 1 chair a row breaks the conditions.
  20. 5 ways: 1 row of 56, 2 rows of 28, 4 rows of 14, 7 rows of 8, 8 rows of 7. The factor pairs are 1 \times 56, 2 \times 28, 4 \times 14, 7 \times 8, and the factors up to 8 are 1, 2, 4, 7, 8.
  21. 4 rows of 11 chairs. The factors of 44 are 1, 2, 4, 11, 22, 44, and only 11 is more than 5 and less than 15. Check: 4 \times 11 = 44.
  22. No. Only square numbers have an odd count, and 7 \times 7 = 49 and 8 \times 8 = 64, so no square number lies from 51 to 59.
  23. Multiplying 7 by the whole number 9 gives 63 exactly, so 63 \div 7 = 9 \text{ R } 0 and 7 is a factor. Turning 7 rows of 9 into 9 rows of 7 (the commutative property) gives 63 \div 9 = 7 \text{ R } 0, so 9 is a factor too. So 63 is divisible by 7 and by 9, and 63 is a multiple of 7 and a multiple of 9.
  24. 25 = 5 \times 5 is a square number, so one of its pairs adds only one factor and the count is odd (it is 1, 5, 25). 26 is not a square, since 5 \times 5 = 25 and 6 \times 6 = 36, so every pair is two different numbers and the count is even (it is 1, 2, 13, 26).
  25. 99 is bigger than 90, but 99 has only 6 factors (1, 3, 9, 11, 33, 99) while 90 has 12. How many factors a number has depends on how many ways it splits into equal rows, not on its size.
kg1128 · updated 2026-10-20 · CC-BY-SA-4.0