 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-4/1e/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Since

4^{- 1} \equiv 16 (mod 21)

and

2^{- 1} \equiv 23 (mod 45)

, the two linear congruences reduce to

x \equiv 16 (mod 21), x \equiv 25 (mod 45) .

Write

x = 16 + 21 k

. The second congruence becomes

21 k \equiv 9 (mod 45),

7 k \equiv 3 (mod 15)

after division by three. Since

7^{- 1} \equiv 13 (mod 15)

k \equiv 9 (mod 15)

. The Chinese remainder theorem therefore gives the complete family

x \equiv 205 (mod 315) .

The modulus is

lcm (21, 45) = 315

because the compatibility condition modulo their greatest common divisor is satisfied.

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