 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ia/paper-4/2d/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The congruences reduce to

z \equiv 14 (mod 17)

and

z \equiv 8 (mod 19)

z = 14 + 17 k

gives

k \equiv 3 (mod 19)

, hence

z \equiv 65 (mod 323)

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