Hard GCSE Prime factorisation Questions

Challenging, exam-style GCSE Prime factorisation questions with worked solutions. Stretch yourself on the hardest Venn method, HCF and LCM from prime factors, prime factorisation, index form problems.

Venn methodHCF and LCM from prime factorsprime factorisationindex formsquare numbersreasoning
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
Explain why prime factorisation gives a reliable method for finding the HCF and LCM even of large numbers, where listing all factors would be slow.
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Worked solution

  1. Every number factorises uniquely

    one prime factorisation\text{one prime factorisation}

    Each whole number has exactly one prime factorisation.

  2. Listing all factors is slow

    many factors\text{many factors}

    A large number can have very many factors.

  3. And error-prone

    easy to miss one\text{easy to miss one}

    It is easy to miss a factor or list one twice.

  4. Prime factorisation is compact

    a few prime powers\text{a few prime powers}

    The factorisation records everything with just a few prime powers.

  5. For the HCF, use shared primes

    lowest powers\text{lowest powers}

    Take each shared prime to its lowest power.

  6. This gives the HCF directly

    HCF\text{HCF}

    No listing of factors is needed.

  7. For the LCM, use highest powers

    highest powers\text{highest powers}

    Take each prime to its highest power.

  8. This gives the LCM directly

    LCM\text{LCM}

    Again, no long list is needed.

  9. You only handle a few numbers

    compare exponents\text{compare exponents}

    You just compare exponents, not dozens of factors.

  10. So it scales to large numbers

    works for big numbers\text{works for big numbers}

    The method stays quick even when the numbers are large.

  11. Example numbers

    84=22×3×784=2^2\times3\times7

    Take 84, for instance.

  12. And another

    360=23×32×5360=2^3\times3^2\times5

    And 360.

  13. HCF quickly

    22×3=122^2\times3=12

    Lowest shared powers give HCF = 12 straight away.

  14. LCM quickly

    23×32×5×7=25202^3\times3^2\times5\times7=2520

    Highest powers give LCM = 2520 without listing factors.

  15. State the point

    unique factorisation\text{unique factorisation}

    Because prime factorisation is unique and compact, it gives a reliable, fast way to find HCF and LCM even for large numbers.

Answer
Prime factorisation is unique, so you compare prime powers (lowest for HCF, highest for LCM) instead of listing every factor — fast and reliable even for large numbers
Question 2
5 markschallenging
Find the HCF and LCM of 8484 and 360360 using their prime factorisations.
Show worked solution

Worked solution

  1. Factorise 84

    84=22×3×784=2^2\times3\times7

    Write 84 as a product of primes.

  2. Factorise 360

    360→180→90→45360\to180\to90\to45

    Halving three times reaches 45.

  3. Finish 360

    45=32×5, ⇒360=23×32×545=3^2\times5,\ \Rightarrow 360=2^3\times3^2\times5

    45=9×545 = 9 \times 5, so 360=23×32×5360 = 2³ \times 3² \times 5.

  4. HCF rule

    common, lowest power\text{common, lowest power}

    Take each shared prime to its lowest power.

  5. Shared primes

    2 and 32\ \text{and}\ 3

    Both share 2 and 3 (7 only in 84, 5 only in 360).

  6. Lowest powers

    22×32^2\times3

    Lowest power of 22 is 222², of 33 is 313¹.

  7. Work out the HCF

    4×3=124\times3=12

    So the HCF is 12.

  8. LCM rule

    each prime, highest power\text{each prime, highest power}

    Take every prime to its highest power.

  9. List the highest powers

    23, 32, 5, 72^3,\ 3^2,\ 5,\ 7

    232³ (from 360360), 323² (from 360360), 55 (from 360360), 77 (from 8484).

  10. Multiply the first two

    8×9=728\times9=72

    23×32=722³ \times 3² = 72.

  11. Multiply by 5

    72×5=36072\times5=360

    72×5=36072 \times 5 = 360.

  12. Multiply by 7

    360×7=2520360\times7=2520

    360×7=2520360 \times 7 = 2520.

  13. So the LCM is

    25202520

    The LCM is 2520.

  14. Check with the product rule

    12×2520=84×36012\times2520=84\times360

    HCF×LCM=30240=84×360\text{HCF} \times \text{LCM} = 30240 = 84 \times 360, confirming.

  15. State the answers

    HCF=12, LCM=2520\text{HCF}=12,\ \text{LCM}=2520

    So HCF = 12 and LCM = 2520.

Answer
HCF = 12, LCM = 2520
Question 3
5 markschallenging
Find the smallest number that is divisible by every whole number from 11 to 66, using prime factorisation.
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Worked solution

  1. What we need

    divisible by 1,2,3,4,5,6\text{divisible by }1,2,3,4,5,6

    We want the smallest number divisible by every whole number from 1 to 6.

  2. That is the LCM of these numbers

    LCM(1,…,6)\text{LCM}(1,\ldots,6)

    The smallest common multiple of them all.

  3. 1 divides everything

    1 ignore1\ \text{ignore}

    1 is a factor of every number, so it does not affect the answer.

  4. Factorise the rest

    2, 3, 4=22, 5, 6=2×32,\ 3,\ 4=2^2,\ 5,\ 6=2\times3

    Write each number in terms of primes.

  5. LCM rule

    each prime, highest power\text{each prime, highest power}

    Take each prime to the highest power appearing.

  6. Highest power of 2

    222^2

    The highest is 222² (from 44).

  7. Highest power of 3

    33

    3 appears to the power 1 (in 3 and 6).

  8. Highest power of 5

    55

    5 appears to the power 1.

  9. Multiply them

    22×3×52^2\times3\times5

    So the LCM is 22×3×52² \times 3 \times 5.

  10. Evaluate

    4×3×54\times3\times5

    22=42² = 4, then multiply by 33 and 55.

  11. Work it out

    4×3×5=604\times3\times5=60

    So the LCM is 60.

  12. Check divisibility

    60÷4=15, 60÷6=1060\div4=15,\ 60\div6=10

    60 divides by 4 and 6 exactly.

  13. Check the rest

    60÷2,3,5 whole60\div2,3,5\ \text{whole}

    60 also divides by 2, 3 and 5 exactly.

  14. So 60 works and is smallest

    6060

    Nothing smaller is divisible by all of 1 to 6.

  15. State the answer

    6060

    So the smallest number divisible by every whole number from 1 to 6 is 60.

Answer
6060
Question 4
5 markschallenging
The product of two numbers is 24×33×522^4\times3^3\times5^2 and their HCF is 22×3×52^2\times3\times5. Find their LCM.
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Worked solution

  1. Use the product rule

    HCF×LCM=product\text{HCF}\times\text{LCM}=\text{product}

    For two numbers, HCF×LCM\text{HCF} \times \text{LCM} equals their product.

  2. So rearrange for the LCM

    LCM=product÷HCF\text{LCM}=\text{product}\div\text{HCF}

    Divide the product by the HCF to get the LCM.

  3. Write the product

    24×33×522^4\times3^3\times5^2

    The product of the two numbers is given in index form.

  4. Write the HCF

    22×3×52^2\times3\times5

    The HCF is also given in index form.

  5. Divide the powers of 2

    24−2=222^{4-2}=2^2

    Subtract the indices: 4−2=24 - 2 = 2.

  6. Divide the powers of 3

    33−1=323^{3-1}=3^2

    Subtract the indices: 3−1=23 - 1 = 2.

  7. Divide the powers of 5

    52−1=55^{2-1}=5

    Subtract the indices: 2−1=12 - 1 = 1.

  8. So the LCM in index form

    22×32×52^2\times3^2\times5

    Putting the pieces together.

  9. Evaluate the powers

    4×9×54\times9\times5

    22=42² = 4 and 32=93² = 9.

  10. Multiply the first two

    4×9=364\times9=36

    4×9=364 \times 9 = 36.

  11. Multiply by 5

    36×5=18036\times5=180

    36×5=18036 \times 5 = 180.

  12. So the LCM is

    180180

    The LCM is 180.

  13. Why dividing works

    subtract indices\text{subtract indices}

    Dividing powers of the same prime subtracts the indices.

  14. Sense check

    HCF×180=product\text{HCF}\times180=\text{product}

    Multiplying the HCF back by 180 returns the original product.

  15. State the answer

    180180

    So the LCM is 180.

Answer
180180
Question 5
5 markschallenging
Find the LCM of 252^5 and 343^4, giving your answer as an ordinary number.
Show worked solution

Worked solution

  1. Look at the two numbers

    25 and 342^5\ \text{and}\ 3^4

    One is a power of 2, the other a power of 3.

  2. They share no prime

    2≠32\ne3

    2 and 3 are different primes, so the numbers share nothing.

  3. So they are coprime

    HCF=1\text{HCF}=1

    Their highest common factor is 1.

  4. For coprime numbers, LCM = product

    LCM=25×34\text{LCM}=2^5\times3^4

    With no shared primes, the LCM is just the product.

  5. Evaluate the first power

    25=322^5=32

    2⁵ = 32.

  6. Evaluate the second power

    34=813^4=81

    3⁴ = 81.

  7. Multiply them

    32×8132\times81

    Now multiply 32 by 81.

  8. Split 81 to make it easier

    32×80+32×132\times80+32\times1

    Break 81 into 80 + 1.

  9. First part

    32×80=256032\times80=2560

    32×80=256032 \times 80 = 2560.

  10. Second part

    32×1=3232\times1=32

    32×1=3232 \times 1 = 32.

  11. Add them

    2560+32=25922560+32=2592

    So 32×81=259232 \times 81 = 2592.

  12. So the LCM is

    25922592

    The LCM is 2592.

  13. Note the HCF

    HCF=1\text{HCF}=1

    As they are coprime, the HCF is 1.

  14. Check with the product rule

    1×2592=32×811\times2592=32\times81

    HCF×LCM=1×2592=2592=product\text{HCF} \times \text{LCM} = 1 \times 2592 = 2592 = \text{product}, which is consistent.

  15. State the answer

    25922592

    So the LCM of 2⁵ and 3⁴ is 2592.

Answer
25922592

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