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!

Solution 1:

In your shoes, I would have tried Wolfram Alpha. Something like solve 21x = 241 mod 1000 should give you the answer:

$x = 821 + 1000n$ and $n \in \mathbb Z$

Well, that needs to be qualified with $n \geq 0$, since negative $n$ give numbers like $-3759$. Positive $n$, and 0, will give you the numbers you want: 17241, 59241, 80241, etc.

Also observe that the last three digits of multiples of 21 cycle at every thousand, which is a bit longer than, say, multiples of 22 (which cycle the last three digits every five hundred multiples).

Solution 2:

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.