Foundation November 2023 Paper 1 Q17
17 A prime number is a whole number that has exactly two factors.
(a) Explain why 1 is not a prime number. [1]
(b) \(a\) and \(b\) are prime numbers.
Write down the 6 factors of \(a^2b\). [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| it has only one factor | 1 | Accept any correct reason see appendix if more than one statement mark the best as long as it is not contradicted or has an incorrect statement | |
Appendix
Exemplar responses for Q17
| Response | Mark |
|---|---|
| It’s a square number and square numbers are not prime | 1 bod |
| Its only factor is 1 | 1 |
| It does not have two factors only one | 1 |
| It can only be \(1 \times 1\) | 1bod |
| It only has itself as a factor | 1 |
| It can only be divided by itself | 1 |
| Doesn’t have exactly 2 factors | 0 |
| It does not have two factors | 0 |
| It’s a square number | 0 |
| It’s only multiples are 1 and itself | 0 |
| It only goes into itself | 0 |
| It only has one prime factor | 0 |
| There is only \(1 \times 1 = 1\) there are no other factors to make 1 and it’s not a whole number (spoilt) | 0 |
| It does not have two factors that are not 1 or 0 | 0 |
| It has the same factor | 0 |
| It does not have any factors | 0 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(1, a, b, ab, a^2, a^2b\) | 2 | B1 for at least 3 correct | Ignore repetitions, maximum 6 values, if more than 6 apply choice, condone e.g \(a \times a\), \(a \times b\) for \(a^2\), \(ab\) etc |