2017 AMC 8 Problems/Problem 21

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 1

There are $2$ cases to consider:

Case $1$ : $2$ of $a$ , $b$ , and $c$ are positive and the other is negative. WLOG, we can 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 $\boxed{\textbf{(A)}\ 0}$ .

Solution 2

Assuming numbers:

WLOG $a=1, b=2,$ and $c=-3$ (Other numbers can apply for $a, b,$ and $c$ as long as their sum is $0$ .) . Then plug $a, b,$ and $c$ into the given equation. The result is always $\boxed{\textbf{(A)}\ 0}$ .

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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All AJHSME/AMC 8 Problems and Solutions

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2017 AMC 8 Problems/Problem 21

Contents

Problem 21

Solution 1

Solution 2

See Also