Exercise · ZW-CHINESE-REMAINDER-THEOREM-009 · ★★☆ Intermediate

A system with \(x\equiv 0 \pmod 6\) and \(x\equiv 1 \pmod 4\). Does it have a solution? Choose the correct statement.

Choose one option.

Show answer
  • D. No solution, because \(x\) must be even and \(x\) odd

Explanation

The first congruence forces \(x\) to be even and the second forces it to be odd, so there is no solution. The options giving a class modulo \(24\) ignore that \(\gcd(6,4)=2\neq 1\).

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Review the concept: Chinese remainder theorem

The Chinese remainder theorem states that a system of linear congruences with pairwise coprime moduli has a unique solution modulo the product of those moduli.

\[x \equiv a_i \pmod{m_i},\ i=1,\dots,k \Rightarrow x \equiv x_0 \pmod{m_1\cdots m_k}\]

More practice problems: chinese remainder theorem

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