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Suppose in chinese reminder theorem all residue are the same. $$x \equiv a \mod m_1\\ x \equiv a \mod m_2\\ x \equiv a \mod m_3\\ ...\\ x \equiv a \mod m_n$$

Does it mean $ x \equiv a \mod lcm(m_1,m_2,..,m_n)$?
If not, in which cases does?
For exampe if
$$x = -1 \mod 2\\ x \equiv -1 \mod 3\\ x \equiv -1 \mod 4\\ x \equiv -1 \mod 5\\ x \equiv -1 \mod 6\\ $$ concludes $x \equiv -1 \mod 60$

aLLex
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    For CRT the modulii must be co-prime. In your example $\gcd(2,4) \neq 1$. So you cannot apply the standard version of CRT. To answer your first question, if all residues are same then (assuming $\gcd(m_i,m_j)=1$ for all $i \neq j$), we will have $x \equiv a \pmod{\prod_{i}m_i}$. – Anurag A Apr 16 '17 at 18:28
  • @AnuragA $x\equiv -1\pmod{4}$ implies $x\equiv -1\pmod{2}$,I don't really think that the $\gcd(m_i,m_j)$ assumption is needed as he is looking for mod $lcm(m_1,m_2,\cdots,m_n)$ – kingW3 Apr 16 '17 at 18:35
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    @kingW3 in a way you are correct. But I think OP is trying to use CRT without considering the conditions and that is why highlighted that. In fact for this problem you don't even need CRT. – Anurag A Apr 16 '17 at 18:40
  • @kingW3 This generalization should be rather unknown (to be honest, I did not know it either), but nice that you pointed it out. – Peter Apr 16 '17 at 18:51
  • @Peter I wouldn't call this a generalization it works in this particular case,because of the generalization for non pairwise coprime because $a\equiv a\pmod{\gcd(2,4)}$.I've mentioned it so the OP doesn't seem to think that $x\equiv a\pmod{lcm(m_1,m_2,\cdots,m_n)}$ is true only if $\gcd(m_i,m_j)=1$ – kingW3 Apr 16 '17 at 19:21

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