# Difference between revisions of "1988 AIME Problems/Problem 4"

## Problem

Suppose that $|x_i| < 1$ for $i = 1, 2, \dots, n$. Suppose further that $|x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n|.$ What is the smallest possible value of $n$?

## Solution

Since $|x_i| < 1$ then

$$|x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n| < n.$$

So $n \ge 20$. We now just need to find an example where $n = 20$: suppose $x_{2k-1} = \frac{19}{20}$ and $x_{2k} = -\frac{19}{20}$; then on the left hand side we have $\left|\frac{19}{20}\right| + \left|-\frac{19}{20}\right| + \dots + \left|-\frac{19}{20}\right| = 20\left(\frac{19}{20}\right) = 19$. On the right hand side, we have $19 + \left|\frac{19}{20} - \frac{19}{20} + \dots - \frac{19}{20}\right| = 19 + 0 = 19$, and so the equation can hold for $n = \boxed{020}$.

## Solution 2 (Motivating solution)

First off, one can test $1,-1,1$ and find that the $LHS$ is $3$ and the RHS is $1.$ Similarly testing $1,-1,-1,1$ yields $4$ on the LHS and $0$ on the RHS. It seems for every negative we gain twice of that negative on the LHS. However, when we test something like $1,-1,-1,-1,1$ we find the LHS to be $5$ and the RHS to be $1.$ What happened? There were more negatives than positives. Why does this mean that the LHS doesn't grow? There aren't enough positives to "cancel out!" Therefore if for every negative we need a positive for it to cancel out to grow. We can make the $LHS$ grow by approximately $2$ every time we choose a negative and a positive, so if we have $20$ numbers, namely $10$ positive and $10$ negative we can obtain the desired answer.