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# Understand the problem

Suppose $a$, $b$, and $c$ are nonzero real numbers, and $a+b+c=0$. What are the possible value(s) for $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}$ ? $\textbf{(A) }0\qquad\textbf{(B) }1\text{ and }-1\qquad\textbf{(C) }2\text{ and }-2\qquad\textbf{(D) }0,2,\text{ and }-2\qquad\textbf{(E) }0,1,\text{ and }-1$

# 2017 AMC 8 Problem 21

Number Theory
Easy
##### Suggested Book
Excursion in Mathematics

Do you really need a hint? Try it first!

Try to think elementarily . That means from the given relation $$a+ b +c =0$$ , what can be the possible signs of the real numbers , $$a, b , \ and \ c$$ accordingly. Then proceed .
There are $2$ cases to consider: Case $1$: 2 of $a$, $b$, and $c$ are positive and the other is negative. With out loss of generality, we can assume that $a$ and $b$ are positive and $c$ (as the relation is symmetric) is negative. In this case, we have that , $$\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}=1+1-1-1=0.$$ Think about the another similar case .
Case $2$ : 2 of $a$, $b$, and $c$ are positive and the other is negative. Here also without loss of generality, we can 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 $\boxed{\textbf{(A)}\ 0}$.

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