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) 18 minutes — Method: the buses leave together again after a number of minutes that is a multiple of both intervals, and the first such time is the lowest common multiple. Working: the multiples of 6 are 6, 12, 18, 24 … and the multiples of 9 are 9, 18, 27 … The first value in both lists is 18, which is 6 × 3 and 9 × 2. Answer: 18 minutes. The distractors: 54 minutes comes from multiplying 6 by 9, which does give a common multiple but not the lowest one; 3 minutes is the highest common factor of 6 and 9 rather than their lowest common multiple; 15 minutes comes from adding the two intervals together.
- (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.
- (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.
- (c) 9 — Method: factors come in pairs that multiply to give the number, so work through the pairs in order; a factor paired with itself is counted only once. Working: the pairs are 1 × 100, 2 × 50, 4 × 25, 5 × 20 and 10 × 10. The first four pairs give eight different factors, and the last pair adds only one more, so the factors are 1, 2, 4, 5, 10, 20, 25, 50 and 100. Answer: 9. The distractors: 10 comes from counting the pair 10 × 10 as two separate factors; 8 comes from leaving 1 out of the list, on the view that 1 is not a proper factor; 4 comes from writing 100 = 2² × 5² and multiplying the two indices together instead of adding 1 to each index first.
- (a) 36 — Method: for the highest common factor, take the LOWER power of each prime that appears in both numbers. Working: for 2, the lower power is 2² (from 108); for 3, the lower power is 3² (from 72), so the highest common factor is 2² × 3² = 4 × 9 = 36. 216 comes from taking the higher power of each prime instead, 2³ × 3³ = 8 × 27 = 216, which gives the lowest common multiple, not the highest common factor. 108 is simply one of the two numbers, not their highest common factor. 6 comes from multiplying the primes without any powers at all, 2 × 3 = 6. Answer: 36.
- (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.
- (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.
- (b) 5 — Method: list the factors of each number and pick the largest value that appears in both lists. Working: the factors of 15 are 1, 3, 5 and 15; the factors of 25 are 1, 5 and 25. The values in both lists are 1 and 5, and the larger of those is 5. Answer: 5. The distractors: 3 comes from choosing a factor of 15 without checking that it also divides 25; 15 comes from assuming that the smaller of the two numbers is always a factor of the larger one; 75 is the lowest common multiple of 15 and 25, given by taking the highest power of each prime instead of the lowest.
- (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.
- (b) 45 — Method: list multiples of each number until one is shared by both, or use 9 = 3² and 15 = 3 × 5, taking the highest power of each prime. Working: multiples of 9 are 9, 18, 27, 36, 45 …; multiples of 15 are 15, 30, 45 …. The lowest multiple in both lists is 45. 135 comes from working out 9 × 15 = 135, the product of the two numbers rather than their lowest common multiple. 3 is the highest common factor of 9 and 15, not the lowest common multiple. 24 comes from working out 9 + 15 = 24, which is not a multiple of either number. Answer: 45.
- (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.
- (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.
- (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.
- (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.
- (a) Yes, because 120 ends in 0 — Method: a whole number divides exactly by 5 when its last digit is 5 or 0, so look at the final digit. Working: the final digit of 120 is 0, so 120 is a multiple of 5; the division confirms it, since 5 × 24 = 120 with nothing left over. Answer: Yes, because 120 ends in 0. The distractors: the option that says yes because 120 is even reaches the right conclusion from the wrong test, since being even is the test for divisibility by 2, and 14 is even but is not a multiple of 5; saying no because 5 does not divide into 12 comes from ignoring the final digit and testing only the leading digits; saying no because the digits add to 3 applies the digit-sum test, which works for 3 and for 9 but not for 5.
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