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.
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Factors, multiples, primes, HCF and LCM worksheet — GCSE Foundation
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- 1.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.
- 2.The number 72 can be written as 2³ × 3², and the number 108 can be written as 2² × 3³. Work out the highest common factor of 72 and 108.
- 3.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.
- 4.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.
- 5.Two lighthouses flash at the start of the same minute. The first lighthouse flashes every 8 minutes and the second flashes every 12 minutes. Work out how many minutes it will be until they next flash together.
- 6.Work out the lowest common multiple of 9 and 15.
- 7.A florist has 60 red roses and 84 white roses. She wants to make identical bunches using all the flowers, with the greatest possible number of bunches. Work out how many red roses will be in each bunch.
- 8.Which statement about the number 91 is correct?
- 9.Write 60 as a product of its prime factors, using index notation.
- 10.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.
- 11.At a bus station one bus leaves every 6 minutes and another leaves every 9 minutes. The two buses leave together at the start of the day. Work out how long it is until they next leave together.
- 12.Is 120 divisible by 5? Give a reason for your answer.
- 13.Work out how many factors 36 has.
- 14.Write down a prime number between 30 and 40.
- 15.Write 12 as a product of its prime factors.
Answer key
- (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: 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) 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.
- (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) 24 — List multiples of 8 and of 12: multiples of 8 are 8, 16, 24, 32; multiples of 12 are 12, 24, 36. The lowest number in both lists is 24, so the lighthouses next flash together after 24 minutes. Multiplying the two numbers together, 8 × 12, gives 96, which double-counts the common factor of 4 shared by 8 and 12. Working out the highest common factor instead of the lowest common multiple gives 4, far too soon a time for both lighthouses to line up again. Adding the two numbers, 8 + 12, gives 20, which is not even a multiple of either 8 or 12. So the lighthouses next flash together after 24 minutes.
- (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) 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) 91 is not prime, because 91 = 7 × 13. — Check 91 for prime factors up to its square root, which is just under 10: 91 ÷ 7 = 13, and both 7 and 13 are prime, so 91 = 7 × 13 and 91 is not a prime number. Checking only 2, 3 and 5 misses that 7 also needs to be tried — 91 is odd, its digits do not sum to a multiple of 3 (9 + 1 = 10), and it does not end in 0 or 5, so those three checks alone wrongly suggest it is prime. Assuming any odd number ending in 1 must be prime ignores that 91 = 7 × 13 is a counterexample. Misapplying the digit-sum test for 3 by miscounting 9 + 1 as a multiple of 3 wrongly concludes 91 is divisible by 3, when the correct digit sum, 10, is not a multiple of 3. So 91 is not prime, because 91 = 7 × 13.
- (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) 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.
- (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.
- (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.
- (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) 31 — Method: a prime number has exactly two factors, 1 and itself, so check each number between 30 and 40 for other factors. Working: 3 × 11 = 33, so 33 is not prime. 2 × 17 = 34, so 34 is not prime. 4 × 9 = 36, so 36 is not prime. 31 has no factors other than 1 and 31, so it is prime. Answer: 31.
- (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.
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