[discia] GCSE Maths › Number › Prime Factors, HCF and LCM › Practise
  1. Either paper 2 marks

    Write \(1404\) as a product of its prime factors.

    Give your answer in index form.

    Show answer
    \(2^{ 2 } \times 3^{ 3 } \times 13\)

    Keep dividing by the smallest prime that goes in.

    Taking out all the factors of 2: \(1404 = 4 \times 351\), which is 2 twos.

    Taking out all the factors of 3: \(351 = 27 \times 13\), which is 3 threes.

    13 is prime, so there is nothing left to break up.

    \(1404 = 2^{ 2 } \times 3^{ 3 } \times 13\)

    • M1method to find prime factors, e.g. \(2\) twos, \(3\) threes and \(13\)
    • A1\(2^{ 2 } \times 3^{ 3 } \times 13\) in index form
  2. Either paper 3 marks

    Express \(224\) as a product of its prime factors, giving your answer in index form.

    Show answer
    \(2^{5} \times 7\)

    Divide by the smallest prime that goes in, and keep dividing until you reach 1. Every divisor must be a prime.

    \(224 \div 2 = 112,\quad 112 \div 2 = 56,\quad 56 \div 2 = 28,\quad 28 \div 2 = 14,\quad 14 \div 2 = 7,\quad 7 \div 7 = 1\)

    The prime factors are \(2 \times 2 \times 2 \times 2 \times 2 \times 7\).

    Collect the repeated primes into powers:

    \(224 = 2^{5} \times 7\)

    • M1a correct start on \(224\): a division by a prime or a factor-tree split into two factors that multiply to \(224\), for example \(224 \div 2\)
    • M1a complete list of prime factors, \(2 \times 2 \times 2 \times 2 \times 2 \times 7\)
    • A1index form \(2^{5} \times 7\)
  3. Either paper 1 mark

    Calculate the lowest common multiple (LCM) of \(9\), \(12\) and \(39\).

    Show answer
    \(468\)

    Write each number as a product of prime factors:

    \(9 = 3^2\), \(12 = 2^2 \times 3\), \(39 = 3 \times 13\)

    The LCM takes the highest power of each prime that appears in any of them:

    \(2^2 \times 3^2 \times 13 = 468\)

  4. Either paper 2 marks

    Show that \(1625\) can be written as a cube number multiplied by a prime number between \(10\) and \(15\).

    Show answer
    \(1625 = 5^3 \times 13\), where \(5^3 = 125\) is a cube number and \(13\) is prime with \(10 < 13 < 15\)

    Look for a cube number that divides \(1625\) exactly, trying the cube numbers \(8\), \(27\), \(64\), \(125\) in turn, and check that what is left is a prime.

    \(1625 \div 125 = 13\)

    So \(1625 = 125 \times 13\)

    \(125 = 5^3\), so \(1625 = 5^3 \times 13\)

    \(13\) is prime, because its only factors are \(1\) and \(13\).

    \(10 < 13 < 15\), so \(13\) is a prime number between \(10\) and \(15\).

    • M1a cube number factor found: \(1625 = 125 \times 13\), or \(5^3 \times 13\)
    • A1states \(13\) is prime and \(10 < 13 < 15\)
  5. Either paper 2 marks

    Two numbers \(m\) and \(n\) are such that

    \(m\) is an even number

    \(n\) is a multiple of 5

    the highest common factor (HCF) of \(m\) and \(n\) is \(9\)

    Write down a possible value for \(m\) and a possible value for \(n\).

    Show answer
    Any pair that meets all three conditions, for example \(m = 18\) and \(n = 45\)

    Both numbers must be multiples of 9, so write \(m = 9 \times a\) and \(n = 9 \times b\).

    The HCF is only 9 if \(a\) and \(b\) share no common factor other than 1.

    Take \(a = 2\) so that \(m = 9 \times 2 = 18\), which is an even number.

    Take \(b = 5\) so that \(n = 9 \times 5 = 45\), which is a multiple of 5.

    Check: 2 and 5 share no common factor other than 1, so the HCF of 18 and 45 is 9.

    • B1a value of \(m\) that is an even number and has HCF \(9\) with the value given for \(n\)
    • B1a value of \(n\) that is a multiple of 5 and has HCF \(9\) with the value given for \(m\)
  6. Either paper 2 marks

    Written as the product of prime factors, \(4312 = 2^3 \times 7^2 \times 11\) and \(6468 = 2^2 \times 3 \times 7^2 \times 11\).

    Find the highest common factor (HCF) of \(4312\) and \(6468\).

    Give your answer as an integer.

    Show answer
    \(2156\)

    \(4312 = 2^3 \times 7^2 \times 11\) and \(6468 = 2^2 \times 3 \times 7^2 \times 11\)

    Take the lower power of each prime. \(3\) is not a factor of 4312, so it is left out.

    \(2^2 \times 7^2 \times 11 = 2156\)

    • M1the lower power of each prime, \(2^2 \times 7^2 \times 11\)
    • A1the integer 2156
  7. Either paper 6 marks

    \(A = 216\) and \(B = 144\)

    (a)

    Express \(A\) as a product of its prime factors. Give your answer in index form.

    (b)

    Work out the highest common factor (HCF) of \(A\) and \(B\).

    (c)

    Work out the lowest common multiple (LCM) of \(A\) and \(B\).

    Show answer
    (a) \(2^{ 3 } \times 3^{ 3 }\)
    (b) \(72\)
    (c) \(432\)

    Divide by 2 until it no longer goes in, then by 3.

    \(216 = 2^{ 3 } \times 3^{ 3 }\)

    \(A = 2^{ 3 } \times 3^{ 3 }\) and \(B = 2^{ 4 } \times 3^{ 2 }\)

    The HCF takes the lower power of each prime.

    \(2^{ 3 } \times 3^{ 2 } = 72\)

    The LCM takes the higher power of each prime.

    \(2^{ 4 } \times 3^{ 3 } = 432\)

    • (a) M1repeated division or factor tree by primes, e.g. dividing by \(2\) and \(3\)
    • (a) A1\(2^{ 3 } \times 3^{ 3 }\)
    • (b) M1lower power of each prime: \(2^{ 3 }\) and \(3^{ 2 }\)
    • (b) A1HCF \(72\)
    • (c) M1higher power of each prime: \(2^{ 4 }\) and \(3^{ 3 }\)
    • (c) A1LCM \(432\)
  8. Either paper 5 marks

    Cupcakes are sold in packs of \(16\).

    Paper cases are sold in packs of \(28\).

    Omar wants to buy exactly the same number of each, with none left over.

    (a)

    Find the smallest number of cupcakes he can buy.

    (b)

    How many packs of each does he buy?

    Show answer
    (a) \(112\)
    (b) 7 packs of cupcakes and 4 packs of paper cases

    The total has to be a multiple of 16 and a multiple of 28, so the smallest total is the lowest common multiple.

    \(16 = 4 \times 4\) and \(28 = 4 \times 7\), and 4 and 7 share no factor other than 1.

    The LCM keeps the shared 4 once and takes both leftovers:

    \(4 \times 4 \times 7 = 112\)

    \(112 \div 16 = 7\) packs of cupcakes

    \(112 \div 28 = 4\) packs of paper cases

    • (a) M1\(16 = 4 \times 4\) and \(28 = 4 \times 7\), or lists of multiples of both
    • (a) M1LCM method, \(4 \times 4 \times 7\), or a common multiple found
    • (a) A1smallest number \(112\)
    • (b) M1divides \(112\) by \(16\) and by \(28\)
    • (b) A1\(7\) packs of cupcakes and \(4\) packs of paper cases
  9. Either paper 5 marks

    A rectangular wall is \(24\) cm wide and \(108\) cm high.

    It is to be covered completely with identical square tiles. No gaps are allowed and no tile may be cut.

    (a)

    Work out the largest possible side length of a tile.

    (b)

    How many of these tiles are needed?

    Show answer
    (a) \(12\ \text{cm}\)
    (b) \(18\)

    The side of a tile has to divide exactly into 24 and into 108, so it is a common factor of the two. The largest one is the HCF.

    \(24 = 12 \times 2\) and \(108 = 12 \times 9\), and 2 and 9 share no factor other than 1.

    HCF \(= 12\), so the largest tile is 12 cm by 12 cm.

    Along the width: \(24 \div 12 = 2\) tiles.

    Up the height: \(108 \div 12 = 9\) tiles.

    \(2 \times 9 = 18\) tiles altogether.

    • (a) P1recognises the HCF of \(24\) and \(108\) is needed
    • (a) M1common factor method: \(24 = 12 \times 2\) and \(108 = 12 \times 9\)
    • (a) A1largest tile side \(12\) cm
    • (b) M1tiles along each side: \(2\) and \(9\)
    • (b) A1\(18\) tiles
  10. Either paper 2 marks

    \(x\) is a whole number greater than \(14\).

    \(x\) leaves a remainder of \(2\) when it is divided by \(3\), by \(4\) and by \(14\).

    Work out the smallest possible value of \(x\), and the next smallest.

    Show answer
    \(86\) and \(170\)

    \(x - 2\) is exactly divisible by \(3\), by \(4\) and by \(14\), so it is a common multiple of the three numbers.

    The lowest common multiple of \(3\), \(4\) and \(14\) is \(84\).

    The smallest value is the lowest common multiple plus the remainder: \(84 + 2 = 86\)

    The next common multiple is \(2 \times 84 = 168\), so the next value is \(168 + 2 = 170\)

    • M1lowest common multiple of \(3\), \(4\) and \(14\) found, \(84\)
    • A1\(86\) and \(170\), each the remainder \(2\) more than a common multiple