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An integer $a$ leaves the same remainder when divided by $3$ and $12$, it also leaves a remainder of $2$ when divided by $4$, what is the remainder when $a$ is divided by 6?

Here's my attempt, would be glad if someone can confirm whether if I'm right in my arguments.

We are told that $a \equiv r \mod 3$ and $a \equiv r \mod 12$.

We know that $a \equiv r \mod 12 \iff a \equiv r \mod 3$ and $a \equiv r \mod 4$. Since we are told that $a$ leaves a remainder of $2$ when divided by $4$, then $a \equiv 2 \mod 4$. So by the iff statement, we have $a \equiv 2 \mod 12$. But, we also know that $a \equiv r \mod 12 \iff a \equiv r \mod 2$ and $a \equiv r \mod 6$, so $a \equiv 2 \mod 6$.

EDIT: I also want to confirm, is the statement $a \equiv b \mod (mn) \iff a \equiv b \mod m$ and $a \equiv b \mod n$ only true for different $m$ and $n$? For example, obviously $a \equiv 2 \mod (2*2) \iff a \equiv 2 \mod 2$ is false.

EDIT 2: After the discussion in the comments that $a \equiv b \mod (mn) \iff a \equiv b \mod m$ and $a \equiv b \mod n$ only when $gcd(m,n)=1$, I realize the last step of my proof isn't correct, since I said "$a \equiv r \mod 12 \iff a \equiv r \mod 2$ and $a \equiv r \mod 6$", but clearly $gcd(2,6)=2 \neq 1$, how do I complete the proof then?

user3184733
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    I would perhaps be careful using modular notation when writing this down... There is nothing in the modular notation saying that when $a \equiv r \pmod 3$, then $0\leq r < 3$. Notational nitpicking aside, I think the proof itself is fine. – Arthur Nov 10 '16 at 15:56
  • I also agree that the proof is fine. – Zoltan Zimboras Nov 10 '16 at 15:58
  • Thanks, just edited my question with related query. Would appreciate clarification. – user3184733 Nov 10 '16 at 15:59
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    For the question in your EDIT you need gcd(m,n)=1 I think. – coffeemath Nov 10 '16 at 16:06
  • @coffeemath Ah, thanks, I can see why that's the case. So that means for $m$ and $n$ that are not coprime, then we have the implication $a \equiv b \mod m$ and $a \equiv b \mod n$ $\implies$ $a \equiv b \mod lcm(m,n)$, right? – user3184733 Nov 10 '16 at 16:13
  • user3184733 I think you're right sbout the more general version (implication only one way, congruence mod lcm(m,n)). It should be easy to show that if true... [Hmm maby its iff, using lcm(m,n) ..] – coffeemath Nov 10 '16 at 16:17
  • I second the notation critique. You should instead write $\ a\bmod 3 = r = a\bmod 12.,$ Confusing operational vs relational $\rm mod$ is one of the biggest sources of errors made by beginners in elementary number theory. So, unlike the above author, I would not call such critiques "nitpicking" since they often reflect fundamental conceptual misunderstandings. – Bill Dubuque Nov 10 '16 at 16:54
  • for your edit, you need either $,\gcd(m,n) = 1,$ or, more generally, replace $,mn,$ by $,{\rm lcm}(m,n).,$ See the constant case of CRT. – Bill Dubuque Nov 10 '16 at 17:02
  • On the difference between $\rm mod$ as a binary operator vs. ternary relation (congruence) see e.g. my answer here or Arturo Magidin's answer here.. – Bill Dubuque Nov 10 '16 at 17:08

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