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==Problem 21==
==Problem==


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


<math>\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</math>
<math>\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</math>
==Video Solution==
https://youtu.be/7an5wU9Q5hk?t=2362


==Solution 1==
==Solution 1==
Line 19: Line 16:


==Video Solution==
==Video Solution==
https://youtu.be/V9wCBTwvIZo - Happytwin
*https://youtu.be/V9wCBTwvIZo
*https://youtu.be/7an5wU9Q5hk?t=2362


==See Also==
==See Also==

Revision as of 13:10, 18 January 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

See Also

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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