Higher June 2019 Paper 6 Q11
11 You are given that \(270 = 3^3 \times 2 \times 5\) and \(177\,147 = 3^{11}\).
(a)
(i) Find the lowest common multiple (LCM) of 270 and 177 147.
Give your answer using power notation and as an ordinary number. [2]
Give your answer using power notation and as an ordinary number. [2]
(ii) Write 177 147 000 000 as a product of its prime factors. [3]
(b) \(3^n = 177\,147 \times 9^5\)
Find the value of \(n\). [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) \(2 \times 3^{11} \times 5\) 1 771 470 | 1 1 | Condone answers switched | |
| (ii) \(2^6 \times 3^{11} \times 5^6\) | 3 | B1 for \(3^{11}\) in answer and M1 for 2 and 5 identified as factors | Accept written in full without indices e.g. in factor tree |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 21 | 3 | M1 for \(3^2\) or \((3^2)^5\) or \(3^{10}\) seen and M1 for 11 + their 10 soi after attempt at converting \(9^5\) to power of 3 Alternative method by trials: 3 marks for answer 21 but M0 for just converting to ordinary number and a wrong trial | M1M1 for answer \(3^{21}\) e.g. M1M1 for \((3^2)^5 = 3^7\) and \(3^{11} \times 3^7 = 3^{18}\) |