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.
Factors, multiples, primes, HCF and LCM worksheet — GCSE Foundation
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- 1.Work out the lowest common multiple of 9 and 15.
- 2.The number 24 can be written as 2³ × 3, and the number 60 can be written as 2² × 3 × 5. Work out the lowest common multiple of 24 and 60.
- 3.Write 200 as a product of its prime factors, using index notation.
- 4.Work out the highest common factor of 20 and 32.
- 5.A rectangular patio measures 90 cm by 120 cm. Ben wants to cover it exactly with identical square tiles, as large as possible, with no tiles cut. Work out the side length of the largest square tile he can use.
- 6.Work out how many factors 36 has.
- 7.Write 12 as a product of its prime factors.
- 8.The highest common factor of two numbers is 4 and their lowest common multiple is 60. One of the numbers is 20. Work out the other number.
- 9.Priya has 24 red beads and 36 blue beads. She makes identical bracelets, using every bead and with none left over. Work out the greatest number of bracelets she can make.
- 10.Work out 2 × 3 × 5 + 1 and decide whether the result is a prime number.
- 11.Is 120 divisible by 5? Give a reason for your answer.
- 12.Work out the highest common factor of 15 and 25.
- 13.The number 36 can be written as 2² × 3², and the number 84 can be written as 2² × 3 × 7. Work out the highest common factor of 36 and 84.
- 14.Tickets for a fairground ride are sold in packs of 6. Tokens for the dodgems are sold in packs of 10. Yusuf wants to buy the smallest number of packs of each so that he ends up with the same number of ride tickets as dodgem tokens. Work out how many ride tickets that is.
- 15.A gardener has 42 tulip bulbs and 56 daffodil bulbs. She plants them in rows, with every row containing the same number of tulip bulbs and the same number of daffodil bulbs, and no bulbs left over. Work out the greatest number of rows she can plant.
Answer key
- (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.
- (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) 2³ × 5² — Method: divide repeatedly by the smallest prime number, then write any repeated prime using a power. Working: 200 ÷ 2 = 100, 100 ÷ 2 = 50, 50 ÷ 2 = 25, 25 ÷ 5 = 5, and 5 is prime, so 200 = 2 × 2 × 2 × 5 × 5, written as 2³ × 5². 2² × 5³ swaps the two powers, giving 4 × 125 = 500, not 200. 2³ × 5 leaves out one of the two 5s, giving 8 × 5 = 40, not 200. 2 × 5³ leaves out two of the three 2s, giving 2 × 125 = 250, not 200. Answer: 2³ × 5².
- (b) 4 — Method: list the factors of each number and compare them; the highest common factor is the largest number that appears in both lists. Working: the factors of 20 are 1, 2, 4, 5, 10, 20; the factors of 32 are 1, 2, 4, 8, 16, 32. The numbers that appear in both lists are 1, 2 and 4, and the largest of these is 4. 2 is a common factor of 20 and 32 but not the largest one. 8 is a factor of 32 but not of 20, since 20 ÷ 8 is not a whole number. 160 is the lowest common multiple of 20 and 32, not their highest common factor. Answer: 4.
- (a) 30 cm — The tile's side length must be a common factor of 90 and 120. The factors of 90 include 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90; the factors of 120 include 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120. The highest number common to both lists is 30, so the largest square tile has a side length of 30 cm. Picking 15 cm, a common factor but not the largest, gives tiles that are smaller than necessary. Picking 10 cm, also a common factor but smaller still, wastes even more of the possible tile size. Working out the lowest common multiple instead of the highest common factor gives 360 cm, a length far bigger than either side of the patio. So the largest square tile Ben can use has a side length of 30 cm.
- (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.
- (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 — Method: for any two numbers, their highest common factor multiplied by their lowest common multiple equals the product of the two numbers. This holds because the HCF collects every prime factor the two numbers share, and the LCM collects every prime factor that appears in either number, so between them they use each prime factor of the two numbers exactly once — the same primes as the product. Working: 4 × 60 = 240, and 240 ÷ 20 = 12. 15 comes from working out 60 ÷ 4 = 15, dividing the wrong pair of numbers. 16 comes from working out 20 − 4 = 16, subtracting the highest common factor instead of using the product rule. 240 is 4 × 60, the product of the highest common factor and the lowest common multiple, left un-divided by 20. Answer: 12.
- (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.
- (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.
- (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.
- (a) 12 — Compare the powers of each prime that appears in both factorisations. In 2² × 3² and 2² × 3 × 7, the prime 2 appears with power 2 in both, and the prime 3 appears with power 2 in one and only power 1 in the other — take the lower power, 3¹. Multiplying the shared primes at their lower powers, 2² × 3, gives 12. Using power 1 for both primes instead of comparing the powers properly, 2 × 3, gives 6, which misses that 2 is common at power 2, not power 1. Multiplying the primes at their higher powers and including 7, which only appears in 84, gives 2² × 3² × 7, which comes to 252 — this is the lowest common multiple, not the highest common factor. Only spotting that 3 is a common prime and overlooking that 2 is common as well gives 3. So the highest common factor of 36 and 84 is 12.
- (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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