Standard +0.3 This is a straightforward application of the Chinese Remainder Theorem with coprime moduli (7 and 41). While it's a Further Maths topic, the question requires only direct application of a standard algorithm with no conceptual complications or extended reasoning, making it slightly easier than average overall.
So \(x \equiv 25 \pmod{7}\) and \(x \equiv 25 \pmod{41}\) \(\Rightarrow x \equiv 25 \pmod{287}\) since \(\gcd(7, 41) = 1\)
A1, M1, A1
Putting the two together OR \(x = 287n + 25, n \in \mathbb{Z}\)
E1
Explanation
OR \(x = 7a + 4 = 41b + 25\) \(\Rightarrow 7(a-3) = 41b \Rightarrow b = 7c\) since \(\gcd(7, 41) = 1\) \(\Rightarrow a = 41c + 3 \Rightarrow x = 7(41c + 3) + 4 = 7(287c + 25)\) i.e. \(x = 287c + 25 (c \in \mathbb{Z})\)
M1, M1, E1, M1, A1
Eliminating either \(a\) or \(b\); Obtaining \(x\) in terms of a single parameter
OR Reciprocal of 41 = 6 (mod 7) is 6 and Reciprocal of 7 (mod 41) is 6. The Chinese Remainder Theorem guarantees a unique solution mod \(7 \times 41\) since \(\gcd(7, 41) = 1\). Solution is \(x = 287n + r, n \in \mathbb{Z}\), where \(r = 4 \times 6 \times \frac{287}{7} + 25 \times 6 \times \frac{287}{41} \pmod{287}\) i.e. \(r = 2034 \pmod{287} = 25\)
M1, A1, E1, M1, A1
Full implementation of the CRT
| $x \equiv 4 \equiv 11, 18, 25, \ldots \pmod{7}$ $\Rightarrow x \equiv 25 \pmod{7}$ | M1 | Considering equivalents to 4 (mod 7) |
| So $x \equiv 25 \pmod{7}$ and $x \equiv 25 \pmod{41}$ $\Rightarrow x \equiv 25 \pmod{287}$ since $\gcd(7, 41) = 1$ | A1, M1, A1 | Putting the two together OR $x = 287n + 25, n \in \mathbb{Z}$ |
| E1 | Explanation |
| **OR** $x = 7a + 4 = 41b + 25$ $\Rightarrow 7(a-3) = 41b \Rightarrow b = 7c$ since $\gcd(7, 41) = 1$ $\Rightarrow a = 41c + 3 \Rightarrow x = 7(41c + 3) + 4 = 7(287c + 25)$ i.e. $x = 287c + 25 (c \in \mathbb{Z})$ | M1, M1, E1, M1, A1 | Eliminating either $a$ or $b$; Obtaining $x$ in terms of a single parameter |
| **OR** Reciprocal of 41 = 6 (mod 7) is 6 and Reciprocal of 7 (mod 41) is 6. The Chinese Remainder Theorem guarantees a unique solution mod $7 \times 41$ since $\gcd(7, 41) = 1$. Solution is $x = 287n + r, n \in \mathbb{Z}$, where $r = 4 \times 6 \times \frac{287}{7} + 25 \times 6 \times \frac{287}{41} \pmod{287}$ i.e. $r = 2034 \pmod{287} = 25$ | M1, A1, E1, M1, A1 | Full implementation of the CRT |
---
1 Determine the solution of the simultaneous linear congruences
$$x \equiv 4 ( \bmod 7 ) , \quad x \equiv 25 ( \bmod 41 ) .$$
\hfill \mbox{\textit{OCR Further Additional Pure 2018 Q1 [5]}}