| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2019 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Number Theory |
| Type | Simultaneous linear congruences |
| Difficulty | Standard +0.3 Part (a) is a routine linear congruence requiring finding the modular inverse of 7 mod 19, which is straightforward using the extended Euclidean algorithm or inspection. Part (b) requires recognizing that gcd(4,6)=2 does not divide (4-3)=1, making the system inconsistent—this is a direct application of the Chinese Remainder Theorem compatibility condition. While this is Further Maths content, both parts are standard textbook exercises requiring only recall of established techniques without novel problem-solving. |
| Spec | 8.02f Single linear congruences: solve ax = b (mod n)8.02h Simultaneous linear congruences: solve up to three |
3
\begin{enumerate}[label=(\alph*)]
\item Solve $7 x \equiv 6 ( \bmod 19 )$.
\item Show that the following simultaneous linear congruences have no solution.
$$x \equiv 3 ( \bmod 4 ) , x \equiv 4 ( \bmod 6 )$$
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2019 Q3 [4]}}