| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2023 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Number Theory |
| Type | Modular arithmetic properties |
| Difficulty | Moderate -0.8 This is a straightforward modular arithmetic question requiring basic division with remainder (part a) and simple application of divisibility properties (part b). The reasoning in part (b) is direct: since 205 ≡ r (mod 7) and 7|(205×8666), students need only recognize that r×8666 ≡ 0 (mod 7), making this easier than average despite being from Further Maths. |
| Spec | 8.02d Division algorithm: a = bq + r uniquely |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | 205 = 7 29 + 2 i.e. q = 29, r = 2 |
| [1] | 1.1 | |
| 1 | (b) | From (a), since 7 does not divide exactly into 205 (and 7 is |
| Answer | Marks | Guidance |
|---|---|---|
| If 7 | (205×8066), then, by Euclid’s Lemma … | M1 |
| Answer | Marks |
|---|---|
| [2] | 2.4 |
| 2.2a | Use of Euclid’s Lemma |
Question 1:
1 | (a) | 205 = 7 29 + 2 i.e. q = 29, r = 2 | B1
[1] | 1.1
1 | (b) | From (a), since 7 does not divide exactly into 205 (and 7 is
prime), 7 must divide 8666
If 7 | (205×8066), then, by Euclid’s Lemma … | M1
A1
[2] | 2.4
2.2a | Use of Euclid’s Lemma
No need to note that this is because r 0
Allow description of the “Euclid’s Lemma” condition
instead
A complete justification with assumption and
conclusion (hcf(7, 205) = 1)1
\begin{enumerate}[label=(\alph*)]
\item Express 205 in the form $7 q + r$ for positive integers $q$ and $r$, with $0 \leqslant r < 7$.
\item Given that $7 \mid ( 205 \times 8666 )$, use the result of part (a) to justify that $7 \mid 8666$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure AS 2023 Q1 [3]}}