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Past exam of the mathematics course of the University of Cambridge / 2012 / ia / Paper 4 / 1D / ii / Solution

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2012 ia Paper 4 1D ii
Created 2026-09-24 Updated 2026-10-07  0 By others on same topic  0 Discussions Create my own version
Write x=3+18k. The second modular congruence becomes 18k≡−1(mod23). The Bezout identity above says that 9 is the inverse of 18 modulo 23, so k≡−9≡14(mod23). Therefore
x=255,x≡255(mod414).​
(1)
The checks are 255=18⋅14+3=23⋅11+2. The Chinese remainder theorem explains uniqueness modulo 18⋅23=414, since the moduli are coprime integers.

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