Why is the common difference of an arithmetic sequence relatively prime to 3?

[Mr Davis 97]

I'm solving a problem, and the solution makes the following statement: "The common difference of the arithmetic sequence 106, 116, 126, ..., 996 is relatively prime to 3. Therefore, given any three consecutive terms, exactly one of them is divisible by 3."

Why is this statement true? Where does it come from? Is it generalizable to other numbers?

 

[fresh_42]

There are exactly three possible remainders. If none of three consecutive numbers is divisible by three, then there is a remainder $r$ that appears twice in the division by three. Say $a = 3p + r$ and $a + b = 3q + r$. Then $b = 3(q-p)$ is divisible by three. The same holds if $a$ and $a+2b$ have the same remainder. Since three doesn't divide the common difference $b$, there have to be all three possible remainders, i.e. exactly one of three consecutive numbers has remainder $0$.