Factor Pairs

They arrive two at a time.

Every rectangle names two factors

Arrange 18 squares in a rectangle. It can be 1 by 18, 2 by 9 or 3 by 6. Each rectangle has two side lengths, and they multiply to make 18: 1 × 18 = 18, 2 × 9 = 18 and 3 × 6 = 18.

Two numbers that multiply to make 18 are called a factor pair of 18. Both numbers in a pair are factors, so the three pairs give 18 six factors: 1, 2, 3, 6, 9 and 18.

1×182×93×6

The three rectangles of 18: 1 by 18, 2 by 9 and 3 by 6. The wider a rectangle is, the shorter it is.

Walk up from 1

To be sure that no factor is missed, try each number in turn, starting from 1. Each number that divides 18 exactly brings its partner with it.

1 divides 18, and its partner is 18. 2 divides 18, and 18 ÷ 2 = 9, so its partner is 9. 3 divides 18, and its partner is 6. 4 does not divide 18, because 18 ÷ 4 = 4 remainder 2. 5 does not either, because 18 ÷ 5 = 3 remainder 3.

Stop there. 5 × 5 = 25, which is already more than 18, so any number from 5 up would need a partner smaller than itself. Every smaller number has been tried, and each partner that exists has been found. The next factor, 6, is already in the list as the partner of 3.

18 = 1 × 181 × 18

18 = 1 × 18: 1 divides 18, so it names its partner 18; 2 factors so far

Widen the rows from 1 until the pairs cross

Widen the rows one square at a time. When the row width divides 18, the squares close into a full rectangle and the width names its partner. When it does not, the last row is short.

A square number

Walk up from 1 for 36. The pairs are 1 × 36, 2 × 18, 3 × 12, 4 × 9 and 6 × 6. In the last pair, 6 is its own partner, so it is counted once, not twice.

That gives 36 nine factors: 1, 2, 3, 4, 6, 9, 12, 18 and 36. A number that is some number times itself, like 36 = 6 × 6, is called a square number, and a square number always has an odd number of factors.

1×362×183×124×96×6

The five rectangles of 36. The last one is a square, 6 by 6: both of its sides are the same factor, 6.

Three slips

Counting the pairs instead of the factors: 18 has 3 pairs, but each pair holds two factors, so 18 has 6 factors.

Leaving out the first pair. 1 × 18 is a pair like any other, so 1 and 18 are factors of 18.

Subtracting to find a partner. The partner of 3 in 18 is 18 ÷ 3 = 6, not 18 − 3 = 15, because a pair multiplies to make the number: 3 × 6 = 18.

Worked example: Factors: Every Equal Team Size for a School Camp, Then the Sizes the Rules Allow

Question At a school camp, 60 students are to be split into teams for the games, with every team the same size and nobody left out. (a) Teams of 1 and a single team of all 60 count too. List every team size that splits the 60 students equally. (b) The camp rules say that each team must have at least 4 students and that there must be at least 3 teams. Which team sizes are still possible?

  1. 1.Find the factors of 60 in pairs, starting from 1: 1 × 60, 2 × 30, 3 × 20, 4 × 15, 5 × 12 and 6 × 10.

    factor pairs of 60factorpartner1×602×303×204×155×126×10
    factor pairs of 601×602×303×204×155×126×10
    Every team size comes with a number of teams, so find the factors of 60 in pairs.

  2. 2.Try 7: 60 ÷ 7 = 8 remainder 4, so 7 is not a factor. The next number, 8, has 8 × 8 = 64, which is more than 60, so any partner of 8 would be smaller than 8 and already found. The list is complete.

    factor pairs of 60factorpartner1×602×303×204×155×126×107 × 8 = 56 and 7 × 9 = 63, so 7 is not a factor8 × 8 = 64 is past 60, so the list is complete
    factor pairs of 601×602×303×204×155×126×107 × 8 = 56 and 7 × 9 = 63, so 7 is not a factor8 × 8 = 64 is past 60, so the list is complete
    7 is not a factor, and 8 × 8 = 64 is more than 60, so the pairs are complete.

  3. 3.(a) The team sizes are the 12 factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.

    factor pairs of 60factorpartner1×602×303×204×155×126×107 × 8 = 56 and 7 × 9 = 63, so 7 is not a factor8 × 8 = 64 is past 60, so the list is complete12 factors: 12 team sizes
    factor pairs of 601×602×303×204×155×126×107 × 8 = 56 and 7 × 9 = 63, so 7 is not a factor8 × 8 = 64 is past 60, so the list is complete12 factors: 12 team sizes
    (a) The 12 factors of 60 are the 12 possible team sizes.

  4. 4.Apply the first rule, at least 4 students in a team. That crosses out team sizes 1, 2 and 3.

    factor pairs of 60factorpartner1×602×303×204×155×126×107 × 8 = 56 and 7 × 9 = 63, so 7 is not a factor8 × 8 = 64 is past 60, so the list is complete12 factors: 12 team sizesat least 4 in a team: 1, 2 and 3 crossed out
    factor pairs of 601×602×303×204×155×126×107 × 8 = 56 and 7 × 9 = 63, so 7 is not a factor8 × 8 = 64 is past 60, so the list is complete12 factors: 12 team sizesat least 4 in a team: 1, 2 and 3 crossed out
    At least 4 students in a team crosses out team sizes 1, 2 and 3.

  5. 5.Apply the second rule, at least 3 teams. The number of teams is the partner in each pair, so team sizes 30 and 60, which give only 2 teams and 1 team, are crossed out.

    factor pairs of 60factorpartner1×602×303×204×155×126×107 × 8 = 56 and 7 × 9 = 63, so 7 is not a factor8 × 8 = 64 is past 60, so the list is complete12 factors: 12 team sizesat least 4 in a team: 1, 2 and 3 crossed outat least 3 teams: 30 (2 teams) and 60 (1 team) crossed out
    factor pairs of 601×602×303×204×155×126×107 × 8 = 56 and 7 × 9 = 63, so 7 is not a factor8 × 8 = 64 is past 60, so the list is complete12 factors: 12 team sizesat least 4 in a team: 1, 2 and 3 crossed outat least 3 teams: 30 (2 teams) and 60 (1 team)crossed out
    At least 3 teams crosses out team sizes 30 and 60, which give 2 teams and 1 team.

  6. 6.(b) The team sizes still possible are 4, 5, 6, 10, 12, 15 and 20, which is 7 sizes. Check: teams of 20 make 60 ÷ 20 = 3 teams, just enough, and teams of 4 make 15 teams.

    factor pairs of 60factorpartner1×602×303×204×155×126×107 × 8 = 56 and 7 × 9 = 63, so 7 is not a factor8 × 8 = 64 is past 60, so the list is complete12 factors: 12 team sizesat least 4 in a team: 1, 2 and 3 crossed outat least 3 teams: 30 (2 teams) and 60 (1 team) crossed out7 sizes left: 4, 5, 6, 10, 12, 15, 20
    factor pairs of 601×602×303×204×155×126×107 × 8 = 56 and 7 × 9 = 63, so 7 is not a factor8 × 8 = 64 is past 60, so the list is complete12 factors: 12 team sizesat least 4 in a team: 1, 2 and 3 crossed outat least 3 teams: 30 (2 teams) and 60 (1 team)crossed out7 sizes left: 4, 5, 6, 10, 12, 15, 20
    (b) Team sizes 4, 5, 6, 10, 12, 15 and 20 are left: 7 sizes.

Answer: (a) 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60 (12 sizes); (b) 4, 5, 6, 10, 12, 15 and 20 (7 sizes)

Common mistakes

  • Leaving out 1 and 60, or stopping the list at 6. Every factor has a partner, so listing the factors in pairs finds 10, 12, 15, 20, 30 and 60 along with 1 to 6.
  • Applying the rule of at least 3 teams to the team size instead of to the number of teams. A team of 30 students is large, but it gives only 60 ÷ 30 = 2 teams, so it is the one that breaks the rule.

More hcf and lcm problems, worked step by step →

Worked example: Factor Pairs: 96 Chairs in a Rectangle, and the Arrangement the Hall Has Room For

Question A caretaker is setting out 96 chairs in a school hall in a rectangle: every row has the same number of chairs, and there are no gaps. (a) Counting 8 rows of 12 and 12 rows of 8 as two different arrangements, and counting a single row of 96 as an arrangement, how many different arrangements are there? (b) The hall has room for no more than 14 chairs in a row and no more than 10 rows. Which arrangement must the caretaker use?

  1. 1.List the factor pairs of 96, testing 1, 2, 3, and so on: 1 × 96, 2 × 48, 3 × 32, 4 × 24, 6 × 16 and 8 × 12. The numbers 5, 7 and 9 leave remainders, and 10 × 10 = 100 is more than 96, so the list is complete.

    factor pairs of 961 × 962 × 483 × 324 × 246 × 168 × 125, 7, 9 leave remainders; 10 × 10 is past 96
    factor pairs of 961 × 962 × 483 × 324 × 246 × 168 × 125, 7, 9 leave remainders; 10 × 10 is past 96
    The factor pairs of 96. 5, 7 and 9 leave remainders, and 10 × 10 = 100 is more than 96, so the list is complete.

  2. 2.(a) Each pair gives two arrangements, because either number can be the number of rows. So there are 6 × 2 = 12 arrangements.

    factor pairs of 961 × 961 row of 9696 rows of 12 × 482 rows of 4848 rows of 23 × 323 rows of 3232 rows of 34 × 244 rows of 2424 rows of 46 × 166 rows of 1616 rows of 68 × 128 rows of 1212 rows of 85, 7, 9 leave remainders; 10 × 10 is past 966 pairs, each either way round: 12 arrangements
    factor pairs of 961 × 961 row of 9696 rows of 12 × 482 rows of 4848 rows of 23 × 323 rows of 3232 rows of 34 × 244 rows of 2424 rows of 46 × 166 rows of 1616 rows of 68 × 128 rows of 1212 rows of 85, 7, 9 leave remainders; 10 × 10 is past 966 pairs, each either way round: 12 arrangements
    (a) Each pair can be set out either way round: 6 × 2 = 12 arrangements.

  3. 3.Apply the first limit, no more than 14 chairs in a row. That rules out rows of 96, 48, 32, 24 and 16 chairs, and leaves 8 rows of 12, 12 rows of 8, 16 rows of 6, 24 rows of 4, 32 rows of 3, 48 rows of 2 and 96 rows of 1.

    factor pairs of 961 × 961 row of 9696 rows of 12 × 482 rows of 4848 rows of 23 × 323 rows of 3232 rows of 34 × 244 rows of 2424 rows of 46 × 166 rows of 1616 rows of 68 × 128 rows of 1212 rows of 85, 7, 9 leave remainders; 10 × 10 is past 966 pairs, each either way round: 12 arrangementsno more than 14 chairs in a row
    factor pairs of 961 × 961 row of 9696 rows of 12 × 482 rows of 4848 rows of 23 × 323 rows of 3232 rows of 34 × 244 rows of 2424 rows of 46 × 166 rows of 1616 rows of 68 × 128 rows of 1212 rows of 85, 7, 9 leave remainders; 10 × 10 is past 966 pairs, each either way round: 12 arrangementsno more than 14 chairs in a row
    No more than 14 chairs in a row crosses out rows of 96, 48, 32, 24 and 16 chairs.

  4. 4.Apply the second limit, no more than 10 rows. Of those seven arrangements, only 8 rows of 12 has 10 rows or fewer.

    factor pairs of 961 × 961 row of 9696 rows of 12 × 482 rows of 4848 rows of 23 × 323 rows of 3232 rows of 34 × 244 rows of 2424 rows of 46 × 166 rows of 1616 rows of 68 × 128 rows of 1212 rows of 85, 7, 9 leave remainders; 10 × 10 is past 966 pairs, each either way round: 12 arrangementsno more than 14 chairs in a rowno more than 10 rows
    factor pairs of 961 × 961 row of 9696 rows of 12 × 482 rows of 4848 rows of 23 × 323 rows of 3232 rows of 34 × 244 rows of 2424 rows of 46 × 166 rows of 1616 rows of 68 × 128 rows of 1212 rows of 85, 7, 9 leave remainders; 10 × 10 is past 966 pairs, each either way round: 12 arrangementsno more than 14 chairs in a rowno more than 10 rows
    No more than 10 rows crosses out 12, 16, 24, 32, 48 and 96 rows. One arrangement is left.

  5. 5.(b) The caretaker must use 8 rows of 12 chairs. Check: 8 × 12 = 96, 12 chairs is within the limit of 14, and 8 rows is within the limit of 10.

    factor pairs of 961 × 961 row of 9696 rows of 12 × 482 rows of 4848 rows of 23 × 323 rows of 3232 rows of 34 × 244 rows of 2424 rows of 46 × 166 rows of 1616 rows of 68 × 128 rows of 1212 rows of 812 chairs in a row8 rows5, 7, 9 leave remainders; 10 × 10 is past 966 pairs, each either way round: 12 arrangementsno more than 14 chairs in a rowno more than 10 rows
    factor pairs of 961 × 961 row of 9696 rows of 12 × 482 rows of 4848 rows of 23 × 323 rows of 3232 rows of 34 × 244 rows of 2424 rows of 46 × 166 rows of 1616 rows of 68 × 128 rows of 1212 rows of 812 chairs in a row8 rows5, 7, 9 leave remainders; 10 × 10 is past 966 pairs, each either way round: 12 arrangementsno more than 14 chairs in a rowno more than 10 rows
    (b) The caretaker must use 8 rows of 12 chairs: 8 × 12 = 96.

Answer: (a) 12 arrangements; (b) 8 rows of 12 chairs

Common mistakes

  • Counting 6 arrangements, one for each factor pair. 8 rows of 12 and 12 rows of 8 look different in the hall, and the question counts them separately, so each pair gives two.
  • Choosing 12 rows of 8 because 8 chairs is within the limit of 14. It has 12 rows, and the hall has room for only 10, so both limits must be checked for every arrangement.

More hcf and lcm problems, worked step by step →

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