TITLE: Describe the multiples of $21$ that ends with $241$
QUESTION [10 upvotes]: 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!

REPLY [10 votes]: 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).