| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2020 |
| Session | November |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Number Theory |
| Type | Linear congruences (single) |
| Difficulty | Moderate -0.8 Part (a) is straightforward arithmetic (247 mod 31 = 30). Part (b) requires finding the multiplicative inverse of 13 mod 31, which is a standard technique but involves either the extended Euclidean algorithm or trial. This is routine Further Maths content with no conceptual difficulty, making it easier than average overall. |
| Spec | 8.02e Finite (modular) arithmetic: integers modulo n8.02f Single linear congruences: solve ax = b (mod n) |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | 30 (mod 31) or –1 (mod 31) |
| [1] | 1.1 | BC No other answer to be accepted |
| Answer | Marks |
|---|---|
| (b) | 13x ≡ 9 ≡ 40 ≡ 71 ≡ … ≡ 195 |
| so x ≡ 15 (mod 31) OR x = 31n + 15 | M1 |
| Answer | Marks |
|---|---|
| A1 | 1.1 |
| Answer | Marks |
|---|---|
| 2.2a | Repeatedly adding 31s |
| Answer | Marks | Guidance |
|---|---|---|
| Alternative method | M1 | Method for finding reciprocal (inverse) of 13 (mod 31) using (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Then 12 × 13x ≡ 12 × 9 | M1 | Multiplication by the reciprocal |
| ⇒ x ≡ 15 (mod 31) | A1 | correct answer |
Question 1:
1 | (a) | 30 (mod 31) or –1 (mod 31) | B1
[1] | 1.1 | BC No other answer to be accepted
Note: 13 × 19 = 247 = 7 × 31 + 30 ≡ 30 (mod 31)
(b) | 13x ≡ 9 ≡ 40 ≡ 71 ≡ … ≡ 195
so x ≡ 15 (mod 31) OR x = 31n + 15 | M1
A1
A1 | 1.1
1.1
2.2a | Repeatedly adding 31s
arriving at a multiple of 13
n ∈ ℤ need not be stated
Alternative method | M1 | Method for finding reciprocal (inverse) of 13 (mod 31) using (a)
13 × 19 ≡ –1 ⇒ 13 × (19×13×19) ≡ 1 so
19×13×19 ≡ 12 is the reciprocal of 13 (mod 31)
Then 12 × 13x ≡ 12 × 9 | M1 | Multiplication by the reciprocal
⇒ x ≡ 15 (mod 31) | A1 | correct answer
[3]1
\begin{enumerate}[label=(\alph*)]
\item Evaluate $13 \times 19$ modulo 31 .
\item Solve the linear congruence $13 x \equiv 9 ( \bmod 31 )$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure AS 2020 Q1 [4]}}