2017 AMC 8 Problems/Problem 21

Revision as of 14:30, 22 November 2017 by Nukelauncher (talk | contribs) (Created page with "There are <math>2</math> cases to consider: Case <math>1</math>: <math>2</math> of <math>a</math>, <math>b</math>, and <math>c</math> are positive and the other is negative....")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

There are $2$ cases to consider:

Case $1$: $2$ of $a$, $b$, and $c$ are positive and the other is negative. WLOG assume that $a$ and $b$ are positive and $c$ is negative. In this case, we have that \[\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}=1+1-1-1=0.\]

Case $2$: $2$ of $a$, $b$, and $c$ are negative and the other is positive. WLOG assume that $a$ and $b$ are negative and $c$ is positive. In this case, we have that \[\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}=-1-1+1+1=0.\]

In both cases, we get that the given expression equals $\boxed{\textbf{(B)}\ 0}$.

~nukelauncher