Question
Find the least positive integer N such that N β‘ 3 (mod
7), N β‘ 4 (mod 9), and N β‘ 5 (mod 8).Solution
ATQ,
From mod 8: N = 8k + 5. Plug into mod 7: 8k + 5 β‘ k + 5 β‘ 3 (mod 7) β k β‘ β2 β‘ 5 (mod 7). So k = 7t + 5 β N = 8(7t + 5) + 5 = 56t + 45. Now enforce mod 9: 56t + 45 β‘ 2t + 0 (mod 9) since 56β‘2, 45β‘0. So 2t β‘ 4 (mod 9) β t β‘ 2 (mod 9). β t = 9s + 2. Least positive: s=0 β t=2 β N = 56Β·2 + 45 = 157.
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