Multiples of 7 & 11 with remainders

If $a\equiv 4 \enspace mod \enspace 7$ and $a\equiv 9\enspace mod\enspace 11$

Then $a= 11k+9$ for any integer k, looking at $(11k+9)\enspace mod\enspace 7 = 4$ leads to the L.H.S. to cycle {2,6,3,0,4,1,5} for k={0,1,2,3,4,5,6} thus $(11k+9)\enspace mod\enspace 7 = 4$ $\Rightarrow k=4+\frac{lcm(7,11)}{11} \ell \enspace\forall\enspace\ell\in \mathbb{N}$ with smallest natural number $a=11\cdot 4+9 = 53$ . $lcm(7,11)=77$ since 7 and 11 are coprime. Thus $a=53+77m \enspace\forall\enspace m \in \mathbb{N}$

$0<a<1000$ $\Rightarrow 0+77\cdot 0 <53+77m<76 + 77 \cdot 12$ $\Rightarrow 0\leq m\leq 12$

which is $12-0+1=13$ solutions.