HMMT November

Algebra Difficulty 4.8 AIME Find the answer

Determine the sum of all distinct real values of $x$ such that $|||\cdots||x|+x|\cdots|+x|+x|=1$ where there are 2017 $x$ 's in the equation.

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Note that $|x+| x||=2 x$ when $x$ is nonnegative, and is equal to 0 otherwise. Thus, when there are 2017 $x$ 's, the expression equals $2017 x$ when $x \geq 0$ and $-x$ otherwise, so the two solutions to the equation are $x=-1$ and $\frac{1}{2017}$ , and their sum is $-\frac{2016}{2017}$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.