| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2018 |
| Session | March |
| Marks | 8 |
| Topic | Number Theory |
| Type | Coprimality proofs |
| Difficulty | Challenging +1.2 This is a Further Maths number theory question requiring the Euclidean algorithm for coprimality and divisibility arguments. Part (i)(a) is a standard HCF proof, part (i)(b) requires algebraic manipulation with divisibility, and part (ii) applies these results systematically. While it demands careful logical reasoning and multiple steps, the techniques are well-established for Further Maths students and the question provides clear scaffolding through its parts. |
| Spec | 8.02d Division algorithm: a = bq + r uniquely8.02i Prime numbers: composites, HCF, coprimality8.02j Divisibility property: a|b and a|c implies a|(bx+cy) |
| Answer | Marks | Guidance |
|---|---|---|
| M1, A1 | 3.1a, 2.2a | Attempted use; |
| [2] |
| Answer | Marks |
|---|---|
| B1 | 3.1a |
| E1 | 2.4 |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | 1.1a | Attempted division (or equivalent algebra) |
| A1 | 2.1 | Must make this conclusion |
| M1 | 3.1a | \(2n+1\) set equal to at least two factors of their numerical 'remainder' (i.e. FT their '13') |
| A1 | 2.2a | All four solutions (and no extras) |
| [4] |
## Part (i)(a)
**Answer:** If $h = \text{hcf}$ then $h \mid [a(3n+2)+b(2n+1)]$
Choosing $a = 2$ and $b = -3$ gives $h \mid 1$ and so $h = 1$
| **M1, A1** | **3.1a, 2.2a** | Attempted use; |
| **[2]** | | |
## Part (i)(b)
**Answer:** $36n^2 + 3n - 14 = (3n+2)(12n-7)$
Since $\text{hcf}(2n+1, 3n+2) = 1$ and $(2n+1) \mid (3n+2)(12n-7)$
it follows, by **Euclid's Lemma**, that $(2n+1) \mid (12n-7)$
| **B1** | **3.1a** | |
| **E1** | **2.4** | |
| **[2]** | | |
## Part (ii)
**Answer:** $\frac{12n-7}{2n+1} = \frac{6(2n+1)-13}{2n+1} = 6 - \frac{13}{2n+1}$
$\Rightarrow (2n+1) \mid 13$
$2n+1 = \pm 1, \pm 13$
$\Rightarrow n = 0, -1, 6, -7$
| **M1** | **1.1a** | Attempted division (or equivalent algebra) |
| **A1** | **2.1** | Must make this conclusion |
| **M1** | **3.1a** | $2n+1$ set equal to at least two factors of their numerical 'remainder' (i.e. FT their '13') |
| **A1** | **2.2a** | All four solutions (and no extras) |
| **[4]** | | |
---6 You are given that $n$ is an integer.
\begin{enumerate}[label=(\roman*)]
\item (a) Show that $\operatorname { hcf } ( 2 n + 1,3 n + 2 ) = 1$.\\
(b) Hence prove that, if $( 2 n + 1 )$ divides $\left( 36 n ^ { 2 } + 3 n - 14 \right)$, then $( 2 n + 1 )$ divides $( 12 n - 7 )$.
\item Use the results of part (i) to find all integers $n$ for which $\frac { 36 n ^ { 2 } + 3 n - 14 } { 2 n + 1 }$ is also an integer.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure AS 2018 Q6 [8]}}