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

Revision as of 21:24, 9 July 2021

Problem

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

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

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

Video Solution

https://youtu.be/FUEHirfk-tw

https://youtu.be/V9wCBTwvIZo - Happytwin

https://youtu.be/7an5wU9Q5hk?t=2362

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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

Revision as of 21:24, 9 July 2021

Contents

Problem

Solution 1

Video Solution

See Also

@@ Line 17: / Line 17: @@
 ==Video Solution==
 https://youtu.be/FUEHirfk-tw
-https://youtu.be/V9wCBTwvIZo
+https://youtu.be/V9wCBTwvIZo - Happytwin
 https://youtu.be/7an5wU9Q5hk?t=2362