2008/09 AHSMC Part 1 Problem 7

Problem 7 of 16

For a real number $t \ge 0$ , $|t| = t$ , and for a real number $t < 0$ , $|t| = -t$ . The number of real numbers $x$ satisfying $|||||x - 1| - 2| - 3| - 4| - 5| = 0$ is

Show solution

Solution

We have $||||x-1|-2|-3|-4| = 5$ . Now $|||x-1|-2|-3| = 4 \pm 5$ . Since the left side can never be equal to $-1$ , we must have $|||x-1|-2|-3| = 9$ . Similarly, $||x-1|-2| = 12$ , $|x-1| = 14$ and $x = 1 \pm 14 = -13$ or $15$ . The answer is (C).

University of Alberta

Answer choices

Solution