Higher November 2022 Paper 2 Q5
5 Written as the product of prime factors,
\(12\,600 = 2^3 \times 3^2 \times 5^2 \times 7\)
and
\(14\,112 = 2^5 \times 3^2 \times 7^2\)
Work out the highest common factor (HCF) of 12 600 and 14 112
Give your answer as an integer. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| At least two of \(2^3\), \(3^2\), 7 selected eg \(2^3 \times 3^2 \times 7\) or 2 2 2 3 3 7 7 or \(2^2 + 3^2 + 7\) or \(2^3 \times 3^2\) or \(2^3 + 7\) or \(3^2 . 7\) | M1 | allow \(2^3\) to be \(2 \times 2 \times 2\) or 8 allow \(3^2\) to be \(3 \times 3\) or 9 allow 7 to be \(7^1\) selection is implied by inclusion in intersection of overlapping circles M0 inclusion of 5 in selection |
| 504 | A1 |
Additional guidance
| \(8 \times 9 \times 7\) | M1 |
| 8, 9, 49 | M1 |
| \(4 + 9 + 7\) | M1 |
| Intersecting circles with eg only 9 and 7 in the intersection | M1 |
| Allow inclusion of 1 for up to M1 eg \(1 \times 2^3 \times 3^2 \times 7\) | M1 |
| \(2^3 \times 3^2 \times 5 \times 7\) | M0 |
| Answer 504 | M1A1 |
| M1 seen with answer the LCM | M1A0 |