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$(5k + 5) \mod 20$

$\gcd(20,5) = 5$

$20$ and $5$ are divisible by $5$ and $1$.

thus the expression gives the value $0$ $2$ times between $1$ and $19$?

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As you have been posting a few questions to this site, I would encourage you to learn how to format them so the math looks nice. See… and/or… – Gerry Myerson Oct 28 '12 at 22:47
up vote 2 down vote accepted

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.

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ah damn k+1 mod 4 – Gladstone Asder Oct 28 '12 at 22:54

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