I'm kind of confused by this because I thought 0 mod 5 = 0, and 0 mod 6 = 0 as well. So what's an integer that is congruent to one but not the other?
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$\begingroup$"$m$ is congruent to $n$ modulo $r$", typically written $$m\equiv n \pmod r,$$ means simply that $m-n$ is divisible by $r$. That is, $$r\mid (m-n).$$ One way to understand this is that you can get from $m$ to $n$ by adding or subtracting $r$ repeatedly.
This usage is different from the one in computer programming, where $\bmod$ is considered a binary operator giving the remainder when one number is divided by another.
$\endgroup$ $\begingroup$Yes, $0 \equiv 0 \pmod 5$ and $0 \equiv 0 \pmod 6$, but the beauty of modular arithmetic are cycles. Namely, that if for example I choose 17 at random, then $17 \equiv 2 \pmod 5$ but $17 \equiv 5 \pmod 6$. Do you see why? We're not concerned with $0$ for this question. We're concerned with some positive and negative integer that yields that result. Another way to think about the question. "What is a positive integer that is divisible by 5, but not divisible by 6? Now, what is a negative integer that is divisible by 5, but not divisible by 6?"
One example is $15$. $15 \equiv 0 \pmod 5$, and $15 \equiv 3 \pmod 6$. For a negative number, we have $-15$. $-15 \equiv 0 \pmod 5$ and $-15 \equiv 3 \pmod 6$. Do you see why that works?
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