Practise Primes, factors, HCF and LCM. 3 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
M1 for a complete method with at most one arithmetic error, e.g. a factor tree or repeated division, or for all the prime factors found (2, 2, 2, 3, 3, 3) but not written as a product
A1 for \(2 \times 2 \times 2 \times 3 \times 3 \times 3\) or \(2^{3} \times 3^{3}\) (factors in any order)
\(A = 2^3 \times 3 \times 5\) and \(B = 2 \times 3^2 \times 7\). Jo says, 'The highest common factor (HCF) of \(A\) and \(B\) is \(2^3 \times 3^2 = 72\).'
(a) Explain why Jo must be wrong.[1]
(b) Find the HCF of \(A\) and \(B\).[1]
(c) Find the lowest common multiple (LCM) of \(A\) and \(B\). Give your answer as an ordinary number.[2]
Show the answer and mark scheme
(a)Answer: 72 is not a factor of \(A\) (\(A = 120\) and 120 ÷ 72 is not a whole number). For the HCF you take the lowest power of each common prime.
C1 for a correct reason, e.g. 72 is not a factor of \(A\) (or of \(B\)), or \(2^3\) is not a factor of \(B\), or \(3^2\) is not a factor of \(A\), or Jo has used the highest powers instead of the lowest
Worked solution: \(A = 120\) and \(B = 126\). The HCF must divide both numbers, but 72 divides neither. Jo used the highest power of each prime; the HCF uses the lowest power of each prime that is common to both.
(b)Answer: 6
B1 for 6 (or \(2 \times 3\))
Worked solution: The common primes are 2 and 3, each to the lowest power: \(2 \times 3 = 6\).
(c)Answer: 2520
M1 for \(2^3 \times 3^2 \times 5 \times 7\) oe
A1 for 2520
Worked solution: Take each prime to its highest power: \(2^3 \times 3^2 \times 5 \times 7 = 8 \times 9 \times 35 = 2520\).
Question 3Hard3 marks
Ali says, 'For every positive integer \(n\), the value of \(n^2 + n + 41\) is a prime number.' Ali works out the value for \(n = 1, 2, 3, 4\) and 5, and gets a prime number each time.
(a) Explain why Ali's checks do not prove his statement.[1]
(b) By choosing a suitable value of \(n\), show that Ali's statement is wrong.[2]
Show the answer and mark scheme
(a)Answer: Checking some values only shows the statement is true for those values; it does not show it is true for every positive integer.
C1 for explaining that a few examples cannot show the statement is true for all positive integers
Worked solution: A proof must work for every value of \(n\). Five examples say nothing about the values that were not checked.
(b)Answer: E.g. \(n = 41\): \(41^2 + 41 + 41 = 41(41 + 1 + 1) = 41 \times 43\), which is not prime.
M1 for choosing a value of \(n\) that gives a non-prime and working out its value or a factorisation, e.g. \(n = 41\): \(41 \times 43\) (= 1763), or \(n = 40\): \(1681 = 41^2\)
C1 for showing the value has a factor other than 1 and itself, with the conclusion that it is not prime
Worked solution: Take \(n = 41\): every term has a factor of 41, \(41^2 + 41 + 41 = 41 \times (41 + 1 + 1) = 41 \times 43 = 1763\). 1763 has factors 41 and 43, so it is not prime and Ali is wrong.