Question

Suppose that $|x_i|\lt1$ for $i=1,$ $2,$ $\ldots,$ $n.$ Suppose further that

$\begin{aligned}&|x_1|+|x_2|+\cdots+|x_n|\\&=19+\bigl|x_1+x_2\\&\qquad+\cdots+x_n\bigr|.\end{aligned}$

What is the smallest possible value of $n?$

Answer

Answer: 20

Concepts:absolute value inequality extremal argument

Difficulty rating: 1920

Small Hint:

Let $P$ be the sum of the positive terms and $N$ the sum of the absolute values of the negative terms

Big Hint:

The left side minus the final absolute value equals $2\min(P,N)$

Solution:

Let $P$ be the sum of the positive $x_i$ and $N$ the sum of the absolute values of the negative $x_i.$ Then $\begin{aligned}P+N-|P-N|&=2\min(P,N)\\&=19,\end{aligned}$ so both $P$ and $N$ are at least $9.5.$ Because every $|x_i|\lt1,$ each sign requires at least $10$ terms, giving $n\geq20.$ Equality is attainable with ten terms equal to $0.95$ and ten equal to $-0.95,$ so the minimum is $20.$

1988 AIME Problem 4

Answer: 20

Difficulty rating: 1920

Small Hint:

Big Hint:

Solution:

Problem 4 in Other Years