What is the smallest solution to $x = 2 \operatorname{mod} (6y+1)$, $x = 3 \operatorname{mod} (12y+1)$, $x= 3 \operatorname{mod} (18y+1)$? I'd like to solve a system of modular equations, but it's a somewhat unusual system. Specifically, what is the smallest natural number $x$ for which $x = 2\pmod{6y+1}$, and $x=3\pmod{12y+1}$, and $1 \pmod{18y+1}$ for some natural number $y$?
If $y=1$, then WolframAlpha tells me that $x=1654$.  But is it possible that a bigger choice of $y$ could yield a smaller value of $x$?  This came up in the context of Godel coding, by the way.
The bases, call them $p,q$ and $r$, are coprime, so try to solve
$$aqr=2\pmod p\\
bpr=3\pmod q\\
cpq=1\pmod r$$
The answer is $aqr+bpr+cpq\pmod{pqr}$
I don't have reputation to comment, so trying to answer here.
To make all 3 modulo co-prime to each other. we take smallest y&gt;0. naturally if we take y=1 then we have modulo as {7,13,19} and they are prime numbers. so, choice of y seems good.

Now the smallest number should be within ${7*13*19=1729}$, so 1654 is the smallest answer you get.
I don't know what could be the technique which gives  higher value of y and smaller value of x.