Difference between revisions of "2017 AMC 8 Problems/Problem 21"

(Solution)
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Case <math>2</math>: <math>2</math> of <math>a</math>, <math>b</math>, and <math>c</math> are negative and the other is positive. WLOG assume that <math>a</math> and <math>b</math> are negative and <math>c</math> is positive. In this case, we have that <cmath>\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}=-1-1+1+1=0.</cmath>
 
Case <math>2</math>: <math>2</math> of <math>a</math>, <math>b</math>, and <math>c</math> are negative and the other is positive. WLOG assume that <math>a</math> and <math>b</math> are negative and <math>c</math> is positive. In this case, we have that <cmath>\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}=-1-1+1+1=0.</cmath>
  
In both cases, we get that the given expression equals <math>\boxed{\textbf{(B)}\ 0}</math>.
+
In both cases, we get that the given expression equals <math>\boxed{\textbf{(A)}\ 0}</math>.
  
 
~nukelauncher
 
~nukelauncher

Revision as of 13:52, 22 November 2017

Problem 21

Suppose $a$, $b$, and $c$ are nonzero real numbers, and $a+b+c=0$. What are the possible value(s) for $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}$?

$\textbf{(A) }0\qquad\textbf{(B) }1\text{ and }-1\qquad\textbf{(C) }2\text{ and }-2\qquad\textbf{(D) }0,2,\text{ and }-2\qquad\textbf{(E) }0,1,\text{ and }-1$

Solution

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 $\boxed{\textbf{(A)}\ 0}$.

~nukelauncher

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