An arithmetic sequence has a common difference, $d$, that is a positive integer and is greater than 1. The sequence includes the three terms 3, 468 and 2018. What is the sum of all of the possible values of $d$ ?

(An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant, called the common difference. For example, $3,5,7,9$ are the first four terms of an arithmetic sequence with common difference 2.)
In an arithmetic sequence with common difference $d$, the difference between any two terms must be divisible by $d$. This is because to get from any term in the sequence to any term later in the sequence, we add the common difference $d$ some number of times.

In the given sequence, this means that $468-3=465$ is a multiple of $d$ and $2018-468=1550$ is a multiple of $d$.

Thus, we want to determine the possible positive common divisors of 465 and 1550 .

We note that $465=5 \cdot 93=3 \cdot 5 \cdot 31$ and that $1550=50 \cdot 31=2 \cdot 5^{2} \cdot 31$.

Therefore, the positive common divisors of 465 and 1550 are 1,5,31,155. (These come from finding the common prime divisors.)

Since $d>1$, the possible values of $d$ are $5,31,155$. The sum of these is $5+31+155=191$.

ANSWER: 191