[discia] GCSE Maths › Number › Factors and Multiples › Practise
  1. Either paper 1 mark

    Circle the factor of \(48\).

    \(5\)   \(144\)   \(18\)   \(13\)   \(8\)

    Show answer
    \(8\)

    A factor of \(48\) divides into \(48\) exactly, with nothing left over.

    \(48 \div 8 = 6\), so \(8\) is a factor of \(48\).

    \(144\) is a multiple of \(48\) (\(48 \times 3 = 144\)). It is bigger than \(48\), so it cannot divide into \(48\).

    \(48 \div 13 = 3\) remainder \(9\), \(48 \div 18 = 2\) remainder \(12\) and \(48 \div 5 = 9\) remainder \(3\), so none of these is a factor.

  2. Either paper 3 marks
    (a)

    Write down the first five multiples of \(15\).

    (b)

    Work out the lowest common multiple (LCM) of \(15\) and \(25\).

    Show answer
    (a) 15, 30, 45, 60, 75
    (b) \(75\)

    Count up in 15s:

    15, 30, 45, 60, 75

    \(15 = 5 \times 3\) and \(25 = 5 \times 5\).

    3 and 5 share no factor other than 1, so the LCM needs the shared 5 once and both leftovers.

    \(5 \times 3 \times 5 = 75\)

    • (b) M1method for the LCM, e.g. \(5 \times 3 \times 5\) or listing multiples of \(15\) and \(25\)
    • (b) A1LCM \(75\)
  3. Either paper 1 mark

    Here is a list of numbers.

    57, 77, 105, 112, 162

    One of these numbers is a multiple of \(21\).

    Which number?

    Show answer
    \(105\)

    A multiple of \(21\) divides by \(21\) exactly.

    \(105 \div 21 = 5\), so 105 is a multiple of \(21\).

    The other numbers are only multiples of a factor of \(21\), such as \(3\), which is not enough. For example \(162 \div 21 = 7\) remainder \(15\), and \(112 \div 21 = 5\) remainder \(7\).

  4. Either paper 3 marks

    Answer both parts.

    (a)

    Write down all the factors of \(55\).

    (b)

    Amara says, “When two multiples of 7 are added, the answer is always a multiple of 14.”

    Give one example to show that she is wrong.

    Show answer
    (a) \(1,\ 5,\ 11,\ 55\)
    (b) \(14 + 35 = 49\), which is not a multiple of \(14\) (any two multiples of 7 with one odd and one even multiplier will do)

    Factors come in pairs that multiply to \(55\):

    \(1 \times 55, 5 \times 11\)

    So the factors of \(55\) are \(1,\ 5,\ 11,\ 55\).

    Write the multiples as \(7 \times a\) and \(7 \times b\). Their sum is \(7(a + b)\).

    This is a multiple of \(14\) only when \(a + b\) is even, so choose one odd and one even multiplier.

    \(14 = 7 \times 2\) and \(35 = 7 \times 5\), and \(14 + 35 = 49\).

    \(49 \div 14 = 3.5\), which is not a whole number, so \(49\) is not a multiple of \(14\) and Amara is wrong.

    • (a) M1factor pairs of \(55\) found: \(1 \times 55, 5 \times 11\)
    • (a) A1all factors \(1,\ 5,\ 11,\ 55\)
  5. Either paper 3 marks

    Here are three statements about whole numbers.

    (a)

    A prime number can be a multiple of 3.

    Give one example to show that this is correct.

    (b)

    The product of two even numbers is always a multiple of 8.

    Give one example to show that this is not correct.

    (c)

    A square number can be a multiple of 8.

    Give one example to show that this is correct.

    Show answer
    (a) Any valid example, for instance: \(3\) is a prime number and \(3 = 3 \times 1\), so it is a multiple of \(3\).
    (b) Any valid example, for instance: \(2 \times 6 = 12\) and \(12 \div 8 = 1.5\), so \(12\) is not a multiple of \(8\); \(2\) and \(6\) are both even.
    (c) Any valid example, for instance: \(16 = 4^2\) and \(16 = 8 \times 2\), so \(16\) is a square number and a multiple of \(8\).

    To show that the statement is correct, find one example for which the statement works.

    The only factors of \(3\) are \(1\) and \(3\), so it is prime, and \(3 \div 3 = 1\), so it is a multiple of \(3\).

    So the statement is correct.

    To show that the statement is not correct, find one example for which the statement does not work.

    Both \(2\) and \(6\) are divisible by \(2\), so both are even, but \(8 \times 1 = 8\) and \(8 \times 2 = 16\), so \(12\) is not a multiple of \(8\).

    So the statement is not correct.

    To show that the statement is correct, find one example for which the statement works.

    \(4 \times 4 = 16\), so \(16\) is a square number, and \(16 \div 8 = 2\), so it is a multiple of \(8\).

    So the statement is correct.

  6. Either paper 1 mark

    Here are four numbers.

    \(8 \qquad 9 \qquad 19 \qquad 36\)

    Which of these numbers has exactly two factors?

    Circle your answer.

    Show answer
    \(19\)

    List the factors of each number.

    \(8\): 1, 2, 4, 8 (4 factors)

    \(9\): 1, 3, 9 (3 factors)

    \(19\): 1, 19 (2 factors)

    \(36\): 1, 2, 3, 4, 6, 9, 12, 18, 36 (9 factors)

    Only \(19\) has exactly two factors: 1, 19.

  7. Either paper 1 mark
    Here are four numbers.
    3273648
    Circle the number that is a multiple of both \(12\) and \(9\).
    Show answer
    \(36\)

    Test each number against both \(9\) and \(12\). A multiple of both numbers is in the times table of each, so it divides exactly by each.

    \(9 \times 4 = 36\) and \(12 \times 3 = 36\), so \(36\) is a multiple of both \(9\) and \(12\).

    \(27\) is a multiple of \(9\) but not of \(12\): the division with \(12\) leaves a remainder of \(3\).

    \(48\) is a multiple of \(12\) but not of \(9\): the division with \(9\) leaves a remainder of \(3\).

    \(3\) is a factor of both numbers, not a multiple of them.

  8. Either paper 2 marks
    Here are five circles arranged in a cross. The vertical line and the horizontal line each have three circles and share the middle one.
    ◯
    ◯ ◯ ◯
    ◯
    Put the numbers \(2,\ 3,\ 4,\ 6,\ 12\) into the circles so that * each line of three numbers multiplies to \(144\) * the total of the horizontal line is one more than the total of the vertical line. Use each number once.
    Show answer
    middle circle \(12\); vertical line \(3\), \(12\), \(4\) (top and bottom in either order); horizontal line \(2\), \(12\), \(6\) (left and right in either order)

    The middle circle is on both lines, so the other four numbers must make two pairs with the same product.

    Put \(12\) in the middle. Each line is \(12\) times its outer pair, so each outer pair must multiply to \(12\), because \(12 \times 12 = 144\).

    The other four numbers pair up as \(3 \times 4 = 12\) and \(2 \times 6 = 12\).

    Both ways round give two lines that multiply to \(144\), so use the totals to choose.

    The lines total \(12 + 3 + 4 = 19\) and \(12 + 2 + 6 = 20\).

    The horizontal line must be one more, so it takes the pair with the larger total: \(2\) and \(6\), total \(20\).

    The vertical line takes \(3\) and \(4\), total \(19\).

    Vertical: \(3\), \(12\), \(4\). Horizontal: \(2\), \(12\), \(6\).

    • M1both lines multiply to \(144\): \(3 \times 12 \times 4\) and \(2 \times 12 \times 6\) with \(12\) in the middle
    • A1the horizontal line totals \(20\) and the vertical line totals \(19\), a difference of 1
  9. Either paper 3 marks

    \(c\) and \(d\) are positive whole numbers.

    \(c\) is even.

    \(d\) is odd.

    Tick a box for each expression to show whether it is even, odd or if you cannot tell.

    (a)

    Is \(\frac{ 2c + 2 }{2}\) even, odd or can you not tell?

    (b)

    Is \(d^2 + 1\) even, odd or can you not tell?

    (c)

    Is \(\frac{ 10d }{2}\) even, odd or can you not tell?

    Show answer
    (a) odd
    (b) even
    (c) odd

    For \(\frac{ 2c + 2 }{2}\): 2 times the even number plus 2, halved, is the even number plus 1, and even + 1 is odd.

    So the answer is odd.

    For \(d^2 + 1\): odd × odd is odd, so the square is odd, and odd + 1 is even.

    So the answer is even.

    For \(\frac{ 10d }{2}\): 10 ÷ 2 = 5, so this is 5 times an odd number, which is odd.

    So the answer is odd.

  10. Either paper 2 marks

    Find the multiple of \(62\) that is nearest to \(524\).

    Show answer
    \(496\)

    Find the multiples of \(62\) either side of \(524\).

    \(8 \times 62 = 496\) and \(8 + 1 = 9\), so \(9 \times 62 = 558\).

    Distance from \(524\) to each:

    \(524 - 496 = 28\) and \(558 - 524 = 34\)

    \(28\) is less than \(34\), so \(524\) is closer to the multiple below it, \(496\).

    • M1multiples either side of \(524\): \(496\) and \(558\)
    • A1\(496\), which is 28 away, against 34 for the other multiple