Confusion over modulus arithmetic

Question

Given: $x\equiv19\ (\textrm{mod}\ 5)$.

Identify which one of the following options gives integers that are valid solutions for $x$ ?

$-19$ and $7$,

$-11$ and $54$,

$7$ and $26$,

$11$ and $26$.

I then tried to emulate that process as follows.

$54 = 10\cdot 5$ with $r=4$

$54\equiv4\ (\textrm{mod}\ 5)$

$7=1\cdot 5 +2$

$2\ (\textrm{mod}\ 5)$

$11 = 2\cdot 5+1$

$1\ (\textrm{mod}\ 5)$

Basically I thought $x= Remainder\ (\textrm{mod}\ 5)$

Answer 1

$x\equiv y\mod n\iff x-y=kn$ for some integer $k$, so for example, $$54-19=35=5k,$$ where $k=7$.

Answer 2

What you show so far is correct:

$54\equiv 4\pmod 5\\7\equiv 2\pmod 5 \\11\equiv 1\pmod 5$

Perhaps what is more important, is that this equivalence is transitive.
$a\equiv x \pmod 5$ and $b \equiv x \pmod 5 \implies a\equiv b \pmod 5$

or $54\equiv 49\equiv 44\pmod 5$

So what is equivalent to 19?

Answer 3

$x \equiv 19 \mod 5$ means $x$ will have the same remainder that $19$ will have. It doesnt mean that $x$ will have $19$ as a remainder. And $19\equiv 4 \mod 5$ so a $x$ will have remainder $4$.

!!BUT!! we aren't babies anymore. Theres nothing magic about the one and only super secret THE remainder. No-one gives a flying crap what the remainder of something is. There's nothing magical about the remainder $4$ that that, $9$ or $14$ or $-1$ or $-6$ wouldn't do just as well.

What's REALLY important is that $x \equiv 19 \mod 5$ is that $x$ and $19$ are equivalent values some multiple of $5$ apart. $x = 19 + 5k$ for some integer (positive, negative, or zero).

And we don't care what $k$ is either. We are interested in the relation $x$ and $19$ have with each other in that some multiple of $5$ apart.

So

....

$x \equiv 19\mod 5$. That means if we subtract a bunch of $5$s we'll, get $x \equiv 4 \mod 5$.

$-19$. if we add a bunch of $5$ s we'll get $-19\equiv 1 \mod 5$ and $1 \not \equiv 4 \mod 5$ so $-19 \not \equiv 19 \mod 5$. That's out.

$7$. If we subtract a bunch of $5$ we'll get $7\equiv 2\not equiv 4 \mod 5$ so $7 \not \equiv 19 \mod 5$.

Now,.... here's a tip. If $a \equiv b \mod 5$ that means we can add or subtract a $5$ to get from one to the other. That means they are different by a multiple of $5$. So if we subtract them $a -b$ that will be $a - b = 5k$ for some integer $k$. And if we add or subtract a bunch of $5$ fro $a-b$ we'll get $0$.

So $a \equiv b \mod 5$ if and only if $a-b \equiv 0 \mod 5$.

$-11$. $-11 - 19 = -30$ and $-30 = 0 \mod 5$. $-11 \equiv 19 \mod 5$. If we add a bunch of $5$s to $-11$ we will get $-11 + 15 = 4$ and $4 + 15 = 19$. So $-11 \equiv 19 \mod 5$.

$54$. Well we can see $54 \equiv 4 \mod 5$ and $19\equiv 4 \mod 5$ so $54\equiv 19 \mod 5$. We can verify this be seed $19 + 35 = 54$ or that $54 - 19 = 35 \equiv 0 \mod 5$.

$-11$ and $54$ is a pair of solutions.