Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-4/6e/a/ii/solution

Because , choose with by Bézout's identity. Then
satisfies and . Any two simultaneous solutions differ by a multiple of both coprime integers and hence by a multiple of . This proves the two-modulus Chinese remainder theorem.
The three congruences reduce to
The first two give ; imposing the last gives

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