How to find $x$ in $\left\lfloor{\frac{x-2000}{4}}\right\rfloor+x\equiv0\pmod 7$

Question

Find all integers $x$ such that $2000\leq x\leq2100$ and $$\left\lfloor{\frac{x-2000}{4}}\right\rfloor+x\equiv0\pmod 7$$

Please, I have no idea how to proceed... any help is really appreciated

Answer 1

HINT:

If $\left\lfloor{\dfrac{x-2000}{4}}\right\rfloor=a$ where $a$ is an integer

$\implies x-2000=4a+c $ where $c$ is an integer and $0\le c\le3$

$\iff x=2000+4a+c$

So, we need $a+2000+4a+c\equiv0\pmod7$

As $2000\equiv5\pmod7,5\equiv-2,$

$ -2a-2-6c\equiv0\iff a\equiv-1-3c\equiv4c-1$

Check for $c=0,1,2,3$