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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

Maths · GCSE Foundation · N4 · Factors, multiples, primes, HCF and LCM · 15 questions

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1.Which statement about the number 91 is correct?
(a)91 is not prime, because 91 = 7 × 13.
(b)91 is prime, because it is not divisible by 2, 3 or 5.
(c)91 is not prime, because it is divisible by 3.
(d)91 is prime, because it is an odd number ending in 1.
2.Write 200 as a product of its prime factors, using index notation.
(a)2 × 5³
(b)2³ × 5
(c)2² × 5³
(d)2³ × 5²
3.For any two whole numbers, the product of the numbers is equal to the product of their highest common factor and their lowest common multiple. The highest common factor of 6 and 8 is 2, and 6 × 8 = 48. Work out the lowest common multiple of 6 and 8.
(a)24
(b)12
(c)48
(d)96
4.Which statement about the number 51 is correct?
(a)51 is prime, because it is not divisible by 2, 3 or 5.
(b)51 is not prime, because it is divisible by 2.
(c)51 is prime, because it is an odd number.
(d)51 is not prime, because 51 = 3 × 17.
5.Work out the highest common factor of 15 and 25.
(a)75
(b)5
(c)15
(d)3
6.A number, n, is a multiple of both 6 and 9. Work out the smallest possible value of n that is greater than 20.
(a)36
(b)18
(c)54
(d)27
7.Work out 2 × 3 × 5 + 1 and decide whether the result is a prime number.
(a)30, which is not prime
(b)31, which is not prime because 31 = 1 × 31
(c)31, which is not prime because it is a multiple of 3
(d)31, which is prime
8.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.
(a)12
(b)7
(c)20
(d)5
9.Work out the highest common factor of 20 and 32.
(a)8
(b)4
(c)2
(d)160
10.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.
(a)36
(b)6
(c)108
(d)216
11.Work out the lowest common multiple of 9 and 15.
(a)24
(b)45
(c)135
(d)3
12.Work out how many factors 100 has.
(a)8
(b)4
(c)9
(d)10
13.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.
(a)7
(b)14
(c)2
(d)168
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.
(a)16
(b)30
(c)60
(d)2
15.Write 90 as a product of its prime factors.
(a)2 × 3² × 5
(b)2 × 45
(c)2 × 3 × 15
(d)3 × 3 × 10

Answer key

(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.
(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.
(a) 24 — Method: rearrange the relationship so that the lowest common multiple stands alone; it is the product of the two numbers divided by their highest common factor. Working: 48 = 2 × the lowest common multiple, so the lowest common multiple is 48 ÷ 2 = 24. Checking, 24 is in the 6 times table and in the 8 times table. Answer: 24. The distractors: 48 comes from giving the product of the two numbers and never dividing by the highest common factor; 96 comes from multiplying by the highest common factor instead of dividing by it; 12 comes from dividing by the highest common factor twice, once for each of the two numbers.
(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.
(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) 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.
(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) 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.
(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) 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.
(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.
(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.
(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.
(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.

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