Discussion
Which of the following is a solution to the system of congruences :
\(x \equiv 2 \pmod{5}\)
\(x \equiv 3 \pmod{4}\)
\(x \equiv 1 \pmod{3}\)
Answer: D
The solution is 67.
Check each candidate against the three congruences:
| x | mod 5 | mod 4 | mod 3 |
|---|---|---|---|
| 22 | 2 | 2 | 1 |
| 27 | 2 | 3 | 0 |
| 47 | 2 | 3 | 2 |
| 67 | 2 | 3 | 1 |
Only 67 gives remainders 2, 3 and 1. By the Chinese remainder theorem the solutions are exactly \(x \equiv 67 \pmod{60}\). 22 fails mod 4, 27 fails mod 3 and 47 fails mod 3.
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