$S=\{1,2,3\ldots,19\}$
$(5k + 5) \mod 20$
$\gcd(20,5) = 5$
$20$ and $5$ are divisible by $5$ and $1$.
thus the expression gives the value $0$ $2$ times between $1$ and $19$?
$S=\{1,2,3\ldots,19\}$
$(5k + 5) \mod 20$
$\gcd(20,5) = 5$
$20$ and $5$ are divisible by $5$ and $1$.
thus the expression gives the value $0$ $2$ times between $1$ and $19$?
This problem is small enough that you can just try all values of $k$, and when you do, you find $k=3$ works (that is, $5k+5\equiv0\pmod{20}$ if $k=3$), and also $k=7$, $k=11$, $k=15$, and $k=19$.
But then you look at those numbers and see they form an arithmetic progression with common difference 4, and you wonder why that should be. And if you think about that long and hard enough, you may learn what's actually going on in this question.