## Question 9:

### Part (i):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Attempts two of $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$, $\overrightarrow{AC} = \mathbf{c} - \mathbf{a}$ and $\overrightarrow{BC} = \mathbf{c} - \mathbf{b}$ either way around | M1 | Condone wrong way around but must be subtraction. Allow marked in correct place on diagram |
| Attempts $\mathbf{c} - \mathbf{b} = 2 \times (\mathbf{b} - \mathbf{a})$, e.g. such as $\mathbf{c} - \mathbf{a} = 3 \times (\mathbf{b} - \mathbf{a})$ | dM1 | Uses the given information. Accept $\overrightarrow{AB} = \frac{1}{3}\overrightarrow{AC}$, $\overrightarrow{BC} = 2 \times \overrightarrow{AB}$, $\overrightarrow{BC} = \frac{2}{3} \times \overrightarrow{AC}$ etc, condoning slips as in previous M1 |
| $\Rightarrow \mathbf{c} = 3\mathbf{b} - 2\mathbf{a}$ | A1* | Fully correct work including bracketing leading to given answer. Expect brackets multiplied out: $\mathbf{c} - \mathbf{b} = 2(\mathbf{b}-\mathbf{a}) \Rightarrow \mathbf{c} - \mathbf{b} = 2\mathbf{b} - 2\mathbf{a} \Rightarrow \mathbf{c} = 3\mathbf{b} - 2\mathbf{a}$ |

**(3 marks)**

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### Part (ii):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Assume there exists a number $n$ that isn't a multiple of 3 yet $n^2$ is a multiple of 3 | B1 | For setting up the contradiction. Minimum: "define a number $n$ such that $n$ is not a multiple of 3 but $n^2$ is" |
| If $n$ is not a multiple of 3 then $m = 3p+1$ or $m = 3p+2$ $(p \in \mathbb{N})$ and attempts to square | M1 | Alternatives: $m = 3p+1$ or $m = 3p-1$. Using modulo 3 arithmetic: $1 \rightarrow 1$ and $2 \rightarrow 4 \equiv 1$ |
| $m^2 = (3p+1)^2 = 9p^2 + 6p + 1$ AND $m^2 = (3p+2)^2 = 9p^2 + 12p + 4 = 3(3p^2+4p+1)+1$ and attempts to square both | M1 A1 | Achieves forms that can be argued as to why they are NOT a multiple of 3. E.g. $m^2=(3p+1)^2=3(3p^2+2p)+1$ or $9p^2+6p+1$ and $m^2=(3p+2)^2=3(3p^2+4p+1)+1$ or $9p^2+12p+4$ |
| $(3p+1)^2 = 9p^2+6p+1\left(=3(3p^2+2p)+1\right)$ is one more than a multiple of 3; $(3p+2)^2 = 9p^2+12p+4$ is not a multiple of 3 as 3 does not divide 4 (exactly) | | Correct reasons required. E.g. $9p^2+12p+4$ is not a multiple of 3 as 4 is not a multiple of 3 |
| Hence if $n$ is a multiple of 3 then $n^2$ is a multiple of 3 | A1 | Minimal conclusion. Correct proof requiring correct calculations, correct reasons, and minimal conclusion. Note B0 M1 M1 M1 A1 is possible |

**(5 marks)**

**Total: 8 marks**