2018 AMC 10B Problem 20

The rule is easy to use, so compute values and look for a pattern.

1. A table. Each value is the previous one, minus the one before that, plus $n$.

For example, $f(3) = 1 - 1 + 3 = 3$ and $f(4) = 3 - 1 + 4 = 6$.
2. Spot the pattern. Compare each value with the one $6$ places earlier: $f(7) = 7 = 1 + 6$, $f(8) = 7 = 1 + 6$, $f(9) = 9 = 3 + 6$, $f(10) = 12 = 6 + 6$, and so on. Each value is $6$ more than the value $6$ steps back:
$$f(n) = f(n-6) + 6.$$
3. Why it keeps going. Write the rule for $n$ and for $n-1$:
$$f(n) = f(n-1) - f(n-2) + n,\qquad f(n-1) = f(n-2) - f(n-3) + (n-1).$$
Put the second into the first. The $f(n-2)$ terms cancel, leaving
$$f(n) = -f(n-3) + 2n - 1.$$
Use this again with $n-3$ in place of $n$: $f(n-3) = -f(n-6) + 2n - 7$. Substituting,
$$f(n) = f(n-6) - 2n + 7 + 2n - 1 = f(n-6) + 6.$$
4. Jump to 2018. $2018 = 2 + 6 \times 336$, so from $f(2)$ we take $336$ jumps of $6$ steps, each adding $6$:
$$f(2018) = f(2) + 6 \times 336 = 1 + 2016 = 2017.$$

The answer is $\boxed{\textbf{(B)}\ 2017}$.

When a rule is simple to apply, a table of the first several values often shows a repeating pattern. Here the pattern repeats every $6$ steps with a steady climb of $6$. Once you have a "jump" rule like that, reach a far-away value by counting jumps from a small value with the same remainder.

Look at the gap $f(n) - n$ in the table: $0, -1, 0, 2, 3, 2$, then again $0, -1, 0, 2, 3, 2$. It repeats every $6$. Since $2018$ leaves remainder $2$ when divided by $6$, it matches $n = 2$, where the gap is $-1$. So $f(2018) = 2018 - 1 = 2017$.