Describe the multiples of $21$ that ends with $241$ Honestly, I have no idea how to proceed in this. I tried to combine both divisibilities critirea of $3$ and $7$, and tried to get a general form for a number being a multiple of $3$ and $7$, but I couldn't get any (and I wasn't hopeful).
If anyone could help me, I would be grateful!

Basically you have
$$x=21a\\x=1000b+241$$
where $a,b,x\in\mathbb N$. It is true for $x=17241,a=821,b=17$.
As I know, there is no known method to obtain all possible solutions if you, in general, replace $21$ and $1000$ with some arbitrary numbers. However, according to the Chinese Remainer Theorem, if the greatest common divisor of the two given numbers is $1$, it then has unique solution modulo $a\cdot b$ (where $a$ and $b$ are the given numbers), as math.h explained in the comment. Therefore, finding one example basically solves the problem.
