# 2018 AMC 10B Problems/Problem 20

## Problem

A function $f$ is defined recursively by $f(1)=f(2)=1$ and $$f(n)=f(n-1)-f(n-2)+n$$for all integers $n \geq 3$. What is $f(2018)$?

$\textbf{(A)} \text{ 2016} \qquad \textbf{(B)} \text{ 2017} \qquad \textbf{(C)} \text{ 2018} \qquad \textbf{(D)} \text{ 2019} \qquad \textbf{(E)} \text{ 2020}$

## Solution 1 (A Bit Bashy)

Start out by listing some terms of the sequence. $$f(1)=1$$ $$f(2)=1$$

$$f(3)=3$$ $$f(4)=6$$ $$f(5)=8$$ $$f(6)=8$$ $$f(7)=7$$ $$f(8)=7$$

$$f(9)=9$$ $$f(10)=12$$ $$f(11)=14$$ $$f(12)=14$$ $$f(13)=13$$ $$f(14)=13$$

$$f(15)=15$$ $$.....$$ Notice that $f(n)=n$ whenever $n$ is an odd multiple of $3$, and the pattern of numbers that follow will always be $+3$, $+2$, $+0$, $-1$, $+0$. The largest odd multiple of $3$ smaller than $2018$ is $2013$, so we have $$f(2013)=2013$$ $$f(2014)=2016$$ $$f(2015)=2018$$ $$f(2016)=2018$$ $$f(2017)=2017$$ $$f(2018)=\boxed{(B) 2017}.$$

## Solution 2 (Bashy Pattern Finding)

Writing out the first few values, we get: $1,1,3,6,8,8,7,7,9,12,14,14,13,13,15,18,20,20,19,19...$. Examining, we see that every number $x$ where $x \equiv 1\pmod 6$ has $f(x)=x$, $f(x+1)=f(x)=x$, and $f(x-1)=f(x-2)=x+1$. The greatest number that's $1\pmod{6}$ and less $2018$ is $2017$, so we have $f(2017)=f(2018)=2017.$ $\boxed B$

## Solution 4 (Algebra)

$$f(n)=f(n-1)-f(n-2)+n.$$ $$f(n-1)=f(n-2)-f(n-3)+n-1.$$ Adding the two equations, we have that $$f(n)=2n-1-f(n-3).$$ Hence, $f(n)+f(n-3)=2n-1$. After plugging in $n-3$ to the equation above and doing some algebra, we have that $f(n)-f(n-6)=6$. Consequently, $$f(2018)-f(2012)=6.$$ $$f(2012)-f(2006)=6.$$ $$\ldots$$ $$f(8)-f(2)=6.$$ Adding these $336$ equations up, we have that $f(2018)-f(2)=6 \cdot 336$ and $f(2018)=\boxed{2017}$.

~AopsUser101