# 2014 UNCO Math Contest II Problems/Problem 2

## Problem

Define the Cheshire Cat function $\fbox{:)}$ by

\begin{align*} \fbox{:)}(x) &= -x \quad \text {if } x \text{ is even and} \\ \fbox{:)}(x) &= x \quad \text{ if }x \text{ is odd} \end{align*}

Find the sum $\fbox{:)}(1) + \fbox{:)}(2) + \fbox{:)}(3) + \fbox{:)}(4) + . . .+ \fbox{:)}(289)$

## Solution

It is evident that we are being asked to find the sum of the sequence $1-2+3-4+...-288+289$.

A straight-forward approach is to find the sum of all the odd numbers from 1 to 289, inclusive, and subtract the sum of all the even numbers from 2 to 288, inclusive. Doing so would give you $145^2 - (144^2 + 144) = (145 - 144)(145 + 144) - 144 = (1)(145 + 144) - 144 = \boxed{145}$.

However, a faster method would be to see that you can group the terms in the sequence $1-2+3-4+...-288+289$ into $1 + (-2 + 3) + (-4 + 5) +...+ (-288 + 289)$. Using this, we can reduce many pairs of positive and negative numbers to $1$s. Since there are as many $1$s as there are odd numbers between 1 and 289, inclusive, and there are 145 odd numbers, the answer is $(1)(145) = \boxed{145}$.