Standard +0.8 This is a straightforward application of the Chinese Remainder Theorem with three moduli that are pairwise coprime. While it requires systematic calculation and careful arithmetic with moderately large numbers (7×37×67=17,353), the method is standard for Further Maths students and involves no conceptual difficulty beyond applying the CRT algorithm. The multi-step arithmetic and potential for calculation errors elevate it slightly above average difficulty.
3 Determine all integers \(x\) for which \(x \equiv 1 ( \bmod 7 )\) and \(x \equiv 22 ( \bmod 37 )\) and \(x \equiv 7 ( \bmod 67 )\).
Give your answer in the form \(\mathrm { x } = \mathrm { qn } + \mathrm { r }\) for integers \(n , q , r\) with \(q > 0\) and \(0 \leqslant \mathrm { r } < \mathrm { q }\).
3 Determine all integers $x$ for which $x \equiv 1 ( \bmod 7 )$ and $x \equiv 22 ( \bmod 37 )$ and $x \equiv 7 ( \bmod 67 )$.\\
Give your answer in the form $\mathrm { x } = \mathrm { qn } + \mathrm { r }$ for integers $n , q , r$ with $q > 0$ and $0 \leqslant \mathrm { r } < \mathrm { q }$.
\hfill \mbox{\textit{OCR Further Additional Pure 2024 Q3 [6]}}