What does x equivalent to 2 mod 15 mean?

Question

I came across the following question:

Consider the following system of equivalences of integers.

$$ x \equiv 2 \bmod{15} $$

$$ x \equiv 4 \bmod{21} $$

The number of solutions in $x$, where $1\le x\le 315$, to the above system of equivalences is

(A) $0$
(B) $1$
(C) $2$
(D) $3$.

Can someone please explain to me what do $x \equiv 2 \bmod{15} $ and $ x \equiv 4 \bmod{21} $ mean?

I know that mod gives the remainder when the former number is divided by the latter like 15 mod 4 = 3.

I need some help about how to solve this sum.

Answer 1

I can give you a nifty formula that will take you straight to the answer, but it doesn't sound like you are ready for that. You need to build up some intuition with these sorts of questions... so try this.

$x \equiv 2\mod15$ would be the same thing as $x = 15n + 2$ where $n$ is an integer. And from the other equation $x = 21m+4$

Now, I leave it to you to do the tough part of finding a combination of $m$ and $n$ that solves both equations.

Answer 2

$$a \equiv b \pmod m \iff m\mid b-a \iff \exists k\ \text{ s.t. }\ mk = b-a$$

The relevant lemma here states: A solution exists iff $2\equiv 4 \pmod{\gcd(15,21)} \iff 2\equiv 4 \pmod 3$. Using the above definition, is this true?

Answer 3

$x\equiv2\pmod{15}$ means that $15$ divides $x-2$
$x\equiv4\pmod{21}$ means that $21$ divides $x-4$

The equations $$ \begin{align} x&\equiv2\pmod{15}&\implies &&x&\equiv2\pmod{3}\\ x&\equiv4\pmod{21}&\implies &&x&\equiv1\pmod{3}\\ \end{align} $$ are inconsistent.