Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $a\equiv b \pmod {m_i}$, $1\leq i\leq k$, there $m_1,m_2,\dots,m_k$ relatively prime, then $a\equiv b\pmod{m_1m_2\cdots m_k}$

My attempt: $$\frac{a-b}{m_i}=t_i, t_i\in Z$$ $$\frac{(a-b)^k}{m_1m_2\cdots m_k}=t_1t_2\cdots t_k$$ $$(a-b)^k\equiv 0 \pmod {m_1m_2\cdots m_k}$$ (not sure about this step) $$a-b\equiv 0 \pmod {m_1m_2\cdots m_k}$$ $$a\equiv b \pmod {m_1m_2\cdots m_k}$$ Is there an error in my proof? I didn't use the fact that $m_1,m_2,\dots,m_k$ are relatively prime and I guess it is given for a reason.

share|cite|improve this question
This is a part of the Chinese Reminder Theorem. – Lior B-S Mar 21 '13 at 9:55

This isn't sure right: $2^k = 0 (mod~4)$ for any $k \ge 2$, but $2\ne 0(mod~4)$.

You can prove your statement easily, note that $$\frac{a-b}{m_i}=t_i, t_i\in Z$$ [$m_i$ are coprime] $$\frac{(a-b)}{m_1m_2\cdots m_k} \in Z$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.