2024 DMC Mock 10 Problems/Problem 12
Revision as of 12:12, 22 December 2024 by Pateywatey (talk | contribs) (Created page with "Take out the absolute values one by one. First, we have <cmath> |||||x-1|-1|-1|-1|-1|=\frac{1}{2} </cmath> so <cmath> ||||x-1|-1|-1|-1|=\frac{1}{2}\text{ or } \frac{3}{2} </...")
Take out the absolute values one by one. First, we have
![\[|||||x-1|-1|-1|-1|-1|=\frac{1}{2}\]](http://latex.artofproblemsolving.com/2/1/4/214c4d000246b7c8be50b932e56669ef50c4195d.png)
so
![\[||||x-1|-1|-1|-1|=\frac{1}{2}\text{ or } \frac{3}{2}\]](http://latex.artofproblemsolving.com/c/7/c/c7cbdf11407593299f67ad109bf65703d22dfc7b.png)
Repeating this process, we eventually get
![\[|x-1| = \frac{1}{2} \text{ or } \frac{3}{2} \text{ or } \frac{5}{2} \text{ or } \frac{7}{2} \text{ or } \frac{9}{2}\]](http://latex.artofproblemsolving.com/d/5/0/d50ea31f5cd7e0469fbbe94ef6db888a3f0e1417.png)
There are two solutions for each value of 
, so there are 
solutions total.
