Higher June 2018 Paper 4 Q3
3
(a)
(i) Write 120 as a product of its prime factors. [3]
(ii) The lowest common multiple (LCM) of \(x\) and 120 is 360.
Find the smallest possible value of \(x\). [2]
(b) Two numbers, A and B, are written as a product of prime factors.\[\text{A} = 2^4 \times 3^2 \times 7^2 \qquad\qquad \text{B} = 2^3 \times 3 \times 5 \times 7\]
Find the highest common factor (HCF) of A and B. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) 2 × 2 × 2 × 3 × 5 oe | 3 | M2 for 2, 2, 2, 3, 5 which could be on a tree diagram or in a table or for an answer one step away from correct answer e.g. 2 × 2 × 2 × 15 or M1 for correct complete method with one error or one step from correct method e.g tree or multiple division or B1 for two of 2, 3 and 5 as factors | So \(2^3 \times 3 \times 5\) scores 3 marks and \(2^3\), 3, 5 scores M2 see additional guidance |
| (ii) 9 | 2 | B1 for an answer of 18, 36, 45, 72, 90, 180 or 360. | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 168 | 2 | M1 for \(2^3\), 3 and 7 selected | condone \(2^3 \times 3 \times 7\) for 2 marks even if calculated incorrectly |