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Mathématiques

Prime Factorisation Practice Questions, With the Square-Root Stopping Rule

Twelve questions on factor pairs, primality testing and factor trees in index form, including why testing whether 97 is prime stops at 9 rather than at 96.

La réponse en bref

A prime number has exactly two factors, itself and 1, which is why 1 is not prime. To write a number as a product of primes, split it repeatedly until every branch is prime: 84 = 2² × 3 × 7. To test whether a number is prime, divide by primes only, up to its square root.

Les exercices

  1. Exercice 1Base2 points

    List all the factors of 36.

    Afficher la correction

    1, 2, 3, 4, 6, 9, 12, 18, 36

    1. Work in pairs from 1 upwards: 1 × 36, 2 × 18, 3 × 12, 4 × 9.
    2. 5 does not divide 36, so no pair there.
    3. 6 × 6 — the pair meets itself at the square root, so stop.
    4. Nine factors in all: 1, 2, 3, 4, 6, 9, 12, 18, 36.
  2. Exercice 2Base2 points

    List all the factors of 48.

    Afficher la correction

    1, 2, 3, 4, 6, 8, 12, 16, 24, 48

    1. Pairs: 1 × 48, 2 × 24, 3 × 16, 4 × 12, 6 × 8.
    2. 5 and 7 do not divide 48.
    3. √48 is about 6.9, so 6 × 8 is the last pair to find.
    4. Ten factors: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.
  3. Exercice 3Base3 points

    Which of 51, 53, 55 and 57 is prime?

    Afficher la correction

    53

    1. 51: digit sum 5 + 1 = 6, divisible by 3, so 51 = 3 × 17.
    2. 55 ends in 5, so 55 = 5 × 11.
    3. 57: digit sum 12, divisible by 3, so 57 = 3 × 19.
    4. 53: not even, digit sum 8, does not end in 0 or 5, and 53 ÷ 7 is not whole. √53 is about 7.3, so the testing is finished — 53 is prime.
  4. Exercice 4Base2 points

    Write 84 as a product of its prime factors in index form.

    Afficher la correction

    2² × 3 × 7

    1. 84 = 2 × 42.
    2. 42 = 2 × 21.
    3. 21 = 3 × 7, and both are prime, so every branch has finished.
    4. 84 = 2 × 2 × 3 × 7 = 2² × 3 × 7.
  5. Exercice 5Standard3 points

    Write 360 as a product of its prime factors in index form.

    Afficher la correction

    2³ × 3² × 5

    1. 360 = 36 × 10.
    2. 36 = 4 × 9 = 2² × 3².
    3. 10 = 2 × 5.
    4. Collecting: 2² × 3² × 2 × 5 = 2³ × 3² × 5.
  6. Exercice 6Standard3 points

    Is 97 prime? Show every division you need to test.

    Afficher la correction

    Yes, 97 is prime

    1. √97 ≈ 9.85, so only primes up to 9 need testing: 2, 3, 5, 7.
    2. 97 is odd, so 2 fails.
    3. Digit sum 9 + 7 = 16, not a multiple of 3, so 3 fails.
    4. 97 does not end in 0 or 5, so 5 fails.
    5. 7 × 13 = 91 and 7 × 14 = 98, so 7 fails. Four tests, not ninety-five — 97 is prime.
  7. Exercice 7Standard3 points

    How many factors does 72 have?

    Afficher la correction

    12

    1. 72 = 8 × 9 = 2³ × 3².
    2. Any factor uses 2 to a power of 0, 1, 2 or 3 — four choices — and 3 to a power of 0, 1 or 2 — three choices.
    3. 4 × 3 = 12 factors.
    4. Check by listing: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72 — twelve of them.
  8. Exercice 8Standard3 points

    Write 2,310 as a product of its prime factors.

    Afficher la correction

    2 × 3 × 5 × 7 × 11

    1. 2,310 is even: 2,310 = 2 × 1,155.
    2. 1,155 has digit sum 12, so 1,155 = 3 × 385.
    3. 385 ends in 5, so 385 = 5 × 77.
    4. 77 = 7 × 11, both prime.
    5. 2,310 = 2 × 3 × 5 × 7 × 11 — five different primes, no index above 1.
  9. Exercice 9Approfondi4 points

    Is 149 prime? Use the square-root rule to decide how far to test.

    Afficher la correction

    Yes, 149 is prime

    1. √149 ≈ 12.2, so test the primes up to 12: 2, 3, 5, 7, 11.
    2. 149 is odd; digit sum 14 is not a multiple of 3; it does not end in 0 or 5.
    3. 7 × 21 = 147, so 7 does not divide it.
    4. 11 × 13 = 143 and 11 × 14 = 154, so 11 does not divide it.
    5. All primes below the square root have failed, so 149 is prime.
  10. Exercice 10Approfondi4 points

    n = 2⁴ × 3² × 5. Find n, and find how many factors it has.

    Afficher la correction

    n = 720, with 30 factors

    1. 2⁴ = 16, 3² = 9, so n = 16 × 9 × 5.
    2. 16 × 9 = 144, and 144 × 5 = 720.
    3. Number of factors: add 1 to each index and multiply.
    4. (4 + 1)(2 + 1)(1 + 1) = 5 × 3 × 2 = 30.
  11. Exercice 11Approfondi4 points

    Find the smallest whole number that 108 must be multiplied by to give a perfect square.

    Afficher la correction

    3, giving 324 = 18²

    1. 108 = 2² × 3³.
    2. A perfect square needs every index to be even.
    3. The index of 2 is already 2; the index of 3 is 3, which is odd.
    4. Multiplying by one more 3 makes it 2² × 3⁴ = 324.
    5. √324 = 18, so 324 is indeed a square.
  12. Exercice 12Approfondi4 points

    Find the smallest whole number k such that 200k is a perfect cube.

    Afficher la correction

    k = 5, giving 1,000 = 10³

    1. 200 = 2³ × 5².
    2. A perfect cube needs every index to be a multiple of 3.
    3. The index of 2 is 3 already; the index of 5 is 2 and needs to reach 3.
    4. So k = 5, giving 2³ × 5³ = 1,000.
    5. 1,000 = 10³, which confirms it.

Où l'on se trompe

  • Calling 1 a prime number, or leaving 1 out of a list of factors.
  • Testing every whole number up to the number itself rather than the primes up to its square root — running 97 against all 95 numbers below it.
  • Stopping a factor tree at a composite branch, writing 84 = 2 × 42 and calling the factorisation finished.
  • Leaving the answer as a list of primes, 2, 2, 3, 7, rather than the product 2² × 3 × 7 in index form.
  • Assuming any odd number is prime, so 51, 57 and 91 get through untested.

Factors come in pairs, and that is where the square root comes from

Every factor of a number has a partner: 36 = 1 × 36 = 2 × 18 = 3 × 12 = 4 × 9 = 6 × 6. Listing them as pairs guarantees none is missed and shows the pairs closing in on each other from both ends until they meet at 6, the square root.

That meeting point is the whole reason the primality test stops where it does. If a number had a factor above its square root, the partner of that factor would be below it, and you would already have found it. So testing 97 means testing 2, 3, 5 and 7 — the primes below its square root of about 9.85 — and nothing else. Four divisions, not ninety-five.

Index form is not tidying up

Writing 360 as 2 × 2 × 2 × 3 × 3 × 5 is correct, and writing it as 2³ × 3² × 5 is the form that is actually usable. Every later technique reads the indices directly: the number of factors is (3 + 1)(2 + 1)(1 + 1) = 24; a perfect square needs every index even; the HCF and LCM of two numbers are built from the smaller and larger of each matching index.

So an answer left as a string of primes has done the work and thrown away the result. The habit worth building is to finish every factor tree by collecting equal primes into powers, in ascending order of the prime.

The two numbers everyone argues about

1 is not prime, because a prime has exactly two distinct factors and 1 has only itself. This is not an arbitrary ruling; if 1 were prime, then 12 = 2² × 3 and 12 = 1 × 2² × 3 and 12 = 1² × 2² × 3 would all be different prime factorisations of the same number, and the uniqueness that makes factorisation useful would be gone.

2 is prime, and it is the only even prime, because every other even number has 2 as a third factor. Odd does not mean prime, and the numbers that catch pupils out are 51 (3 × 17), 57 (3 × 19) and 91 (7 × 13). There are 25 primes below 100; in England, pupils are expected to recall those up to 19 and to be able to establish whether any number up to 100 is prime.

Prime numbers below 100
25Prime numbers below 100
Divisions needed to prove 97 prime, since its square root is about 9.85
4Divisions needed to prove 97 prime, since its square root is about 9.85

Questions fréquentes

Why do you only test up to the square root?

Because factors come in pairs whose product is the number. If a factor larger than the square root existed, its partner would be smaller than the square root, and the test would already have found it. For 97 that means four divisions rather than ninety-five, and the saving grows quickly with larger numbers.

Why is 1 not a prime number?

A prime has exactly two distinct factors, and 1 has only one. The deeper reason is uniqueness: if 1 counted as prime then 12 could be written as 2² × 3, or 1 × 2² × 3, or 1⁵ × 2² × 3, and prime factorisation would stop giving one answer per number. Every technique built on it depends on that.

Does the shape of the factor tree matter?

No. Starting 360 as 36 × 10 or as 4 × 90 or as 2 × 180 leads to the same set of primes every time. That is the fundamental theorem of arithmetic at work, and it is worth letting a child prove it to themselves once by drawing two different trees for the same number.

Which primes should a child know by heart?

The primes up to 19 — 2, 3, 5, 7, 11, 13, 17, 19 — are the ones the English curriculum expects to be recalled, and they are also the only ones needed to test any number below 400. The rest is method: divisibility by 2, 3 and 5 by inspection, then trial division up to the square root.

Sources

  1. National curriculum in England: mathematics programmes of study — Department for Education

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