Printable · GCSE Foundation · ages 14-16
Factors, multiples, primes, HCF and LCM worksheet — GCSE Foundation
Fifteen questions on "factors, multiples, primes, hcf and lcm" — DfE statement N4. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Answer key: Factors, multiples, primes, HCF and LCM worksheet — GCSE Foundation
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- (a) 2² × 3 — Method: divide repeatedly by the smallest prime that goes in, until 1 is reached, then write the primes used as a product with indices. Working: 12 ÷ 2 = 6, 6 ÷ 2 = 3 and 3 ÷ 3 = 1, so the primes used are 2, 2 and 3, which is written as 2² × 3. Answer: 2² × 3. The distractors: 2 × 6 comes from stopping at the first factor pair without splitting the 6, which is not prime; 2 × 3 comes from listing each prime once and losing the repeat, and it multiplies to 6 rather than 12; 2 × 3² puts the index on the wrong prime and multiplies to 18.
- (a) 12 — List the multiples of each number: multiples of 4 are 4, 8, 12, 16, 20, 24; multiples of 6 are 6, 12, 18, 24. The lowest number that appears in both lists is 12. Picking 24, a common multiple but not the lowest one, gives an answer that is too big. Picking 6, the larger of the two original numbers rather than a common multiple, ignores that the lowest common multiple must appear in both lists. Working out the highest common factor instead of the lowest common multiple gives 2. So the lowest common multiple of 4 and 6 is 12.
- (d) 5 — Method: the greatest number of identical bunches is the highest common factor of the two flower totals; then divide the red roses by that number of bunches. Working: 60 = 2² × 3 × 5 and 84 = 2² × 3 × 7, so their highest common factor is 2² × 3 = 12. That means 12 bunches, and 60 ÷ 12 = 5 red roses in each. 7 is the number of white roses in each bunch, since 84 ÷ 12 = 7, not red roses. 12 is the number of bunches itself, not the number of red roses in one bunch. 20 comes from working out 60 ÷ 3 = 20, dividing by only part of the highest common factor. Answer: 5.
- (a) 120 — For the lowest common multiple, take each prime that appears in either factorisation, raised to the higher power. In 2³ × 3 and 2² × 3 × 5, the prime 2 appears with power 3 in one and power 2 in the other — take the higher, 2³; the prime 3 appears with the same power in both, 3¹; and the prime 5 appears only in the second factorisation, so use 5¹. Multiplying these, 2³ × 3 × 5, gives 120. Taking the lower power of 2 instead of the higher, and leaving out 5 altogether, gives the highest common factor, 12, instead. Multiplying the two original numbers together, 24 × 60, gives 1440, which double-counts every shared prime factor. Assuming the lowest common multiple is simply the larger of the two numbers gives 60, but 60 is not a multiple of 24 — 60 ÷ 24 does not divide exactly. So the lowest common multiple of 24 and 60 is 120.
- (d) 12 — Method: if the bracelets are identical and no beads are left over, the number of bracelets must divide exactly into both totals, so it is the highest common factor of 24 and 36. Working: 24 = 2³ × 3 and 36 = 2² × 3²; taking the lower index of each shared prime gives 2² × 3 = 4 × 3 = 12. Each bracelet then has 2 red beads and 3 blue beads. Answer: 12. The distractors: 6 comes from taking each shared prime once rather than at its lower index, giving 2 × 3, which is a common factor but not the highest; 72 is the lowest common multiple of 24 and 36, from taking the higher index of each prime instead of the lower; 60 comes from adding the two bead totals instead of looking for a common factor.
- (d) 31, which is prime — Method: work out the value, remembering that multiplication comes before addition, then test it for primality by dividing by each prime up to its square root. Working: 2 × 3 × 5 = 30, so the value is 30 + 1 = 31. Since 6² = 36 is larger than 31, only 2, 3 and 5 need testing: 31 is odd, 31 ÷ 3 leaves a remainder of 1, and 31 does not end in 0 or 5. It therefore has exactly two factors, 1 and itself. Answer: 31, which is prime. The distractors: 30, which is not prime comes from working out 2 × 3 × 5 and forgetting to add the 1; the claim that 31 = 1 × 31 makes it non-prime comes from treating any factor pair as proof, forgetting that a prime is allowed the pair 1 and itself; the claim that 31 is a multiple of 3 comes from assuming that a number containing the digit 3 divides by 3, when in fact 31 ÷ 3 leaves a remainder.
- (d) 51 is not prime, because 51 = 3 × 17. — Check 51 for small prime factors: 51 ÷ 3 = 17, and both 3 and 17 are themselves prime, so 51 = 3 × 17 and 51 is not a prime number — it has factors other than 1 and itself. Checking only 2, 3 and 5 and concluding wrongly that none of them divide 51 misses that 3 does divide it exactly, so the claim that 51 is prime because it avoids 2, 3 and 5 is false. Assuming any odd number must be prime ignores that 51 = 3 × 17 is a counterexample — plenty of odd numbers are not prime. Misreading 51 as the even number 52 leads to the false claim that it is divisible by 2; 51 itself is odd, and 2 is not one of its factors. So 51 is not prime, because 51 = 3 × 17.
- (a) 2 × 3² × 5 — Method: divide repeatedly by the smallest prime number until only prime factors remain. Working: 90 ÷ 2 = 45, 45 ÷ 3 = 15, 15 ÷ 3 = 5, and 5 is prime, so 90 = 2 × 3 × 3 × 5, written as 2 × 3² × 5. 2 × 3 × 15 stops before the 15 is broken down into 3 × 5, so it is not fully factorised. 3 × 3 × 10 stops before the 10 is broken down into 2 × 5. 2 × 45 stops after only one division. Answer: 2 × 3² × 5.
- (b) 6 — List the factors of each number: the factors of 12 are 1, 2, 3, 4, 6 and 12; the factors of 18 are 1, 2, 3, 6, 9 and 18. The common factors are 1, 2, 3 and 6, and the highest of these is 6. Picking 2, a common factor but not the largest, gives an answer that is too small. Picking 3, also a common factor but still not the largest, gives another answer that is too small. Working out the lowest common multiple instead of the highest common factor gives 36. So the highest common factor of 12 and 18 is 6.
- (a) 36 — Method: find the lowest common multiple of 6 and 9, then move up the list of common multiples until one is greater than 20. Working: the common multiples of 6 and 9 are 18, 36, 54 …. 18 is not greater than 20, so the next one, 36, is the smallest value of n that is greater than 20. 18 is the lowest common multiple itself, but it fails the 'greater than 20' condition. 54 is the common multiple after 36, one step too far. 27 is a multiple of 9 but not of 6, since 27 ÷ 6 is not a whole number. Answer: 36.
- (b) 2² × 3 × 5 — Repeatedly divide 60 by prime numbers: 60 ÷ 2 = 30, 30 ÷ 2 = 15, 15 ÷ 3 = 5, and 5 is itself prime. So 60 is 2 × 2 × 3 × 5, which in index notation is 2² × 3 × 5. Stopping the factor tree after only three divisions and writing 2 × 3 × 5 misses that the 2 divides in twice, and gives only 30, not 60. Squaring the 3 as well as the 2 gives 2² × 3² × 5, which comes to 180, far too big. Squaring the 5 instead of the 2 gives 2 × 3 × 5², which comes to 150, also too big. So 60 = 2² × 3 × 5.
- (a) 6 — Method: list the factors of each number and compare them. Working: the factors of 18 are 1, 2, 3, 6, 9, 18; the factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. The only option that appears in both lists is 6. 8 is a factor of 24 but not of 18. 9 is a factor of 18 but not of 24. 12 is a factor of 24 but not of 18. Answer: 6.
- (d) 9 — List all the factors of 36 in pairs that multiply to give 36: 1 × 36, 2 × 18, 3 × 12, 4 × 9, and 6 × 6. This gives the factors 1, 2, 3, 4, 6, 9, 12, 18 and 36 — nine factors in total, with 6 counted only once even though it appears in a pair with itself. Forgetting that 36 is itself a factor of 36 and leaving it off the list gives 8. Counting the number of factor pairs, five of them, rather than the number of individual factors gives 5. Treating the repeated pair 6 × 6 as two separate factors, 6 and 6 again, gives 10 instead of 9. So 36 has 9 factors.
- (b) 30 — Method: the smallest matching total is the lowest common multiple of the two pack sizes. Working: multiples of 6 are 6, 12, 18, 24, 30 …; multiples of 10 are 10, 20, 30 …. The lowest common multiple is 30. 60 comes from working out 6 × 10 = 60, the product of the pack sizes rather than their lowest common multiple. 16 comes from working out 6 + 10 = 16, which is not a common multiple at all. 2 is the highest common factor of 6 and 10, not a total of tickets. Answer: 30.
- (b) 14 — Method: the greatest number of identical rows is the highest common factor of the two bulb totals, found by taking every prime factor the two totals share. Working: 42 = 2 × 3 × 7 and 56 = 2 × 2 × 2 × 7, so the prime factors common to both are 2 and 7, giving a highest common factor of 2 × 7 = 14. 2 comes from taking only the common factor 2 and forgetting the common factor 7. 7 comes from taking only the common factor 7 and forgetting the common factor 2. 168 is the lowest common multiple of 42 and 56, not their highest common factor. Answer: 14.
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