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

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==Problem 21==
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==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>?
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There are <math>2</math> cases to consider:
 
There are <math>2</math> cases to consider:
  
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Case <math>1</math>: <math>2</math> of <math>a</math>, <math>b</math>, and <math>c</math> are positive and the other is negative. WLOG, we can assume that <math>a</math> and <math>b</math> are positive and <math>c</math> is negative. 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>
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Case <math>1</math>: <math>2</math> of <math>a</math>, <math>b</math>, and <math>c</math> are positive and the other is negative. Without loss of generality (WLOG), we can assume that <math>a</math> and <math>b</math> are positive and <math>c</math> is negative. 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, we can 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, we can 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>
  
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In both cases, we get that the given expression equals <math>\boxed{\textbf{(A)}\ 0}</math>.
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Note these are the only valid cases, for neither <math>3</math> negatives nor <math>3</math> positives would work as they cannot sum up to <math>0</math>. In both cases, we get that the given expression equals <math>\boxed{\textbf{(A)}\ 0}</math>.
  
 +
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
 +
https://youtu.be/wPVQwzA-2hU
 +
 +
~Education, the Study of Everything
 +
 +
==Video Solution by OmegaLearn==
 +
https://youtu.be/7an5wU9Q5hk?t=2362
 +
 +
==Video Solutions==
 +
https://youtu.be/FUEHirfk-tw
 +
 +
https://youtu.be/V9wCBTwvIZo
 +
 +
- Happytwin
 +
 +
https://youtu.be/xN0dnJC1hv8
 +
 +
~savannahsolver
 +
 +
https://www.youtube.com/watch?v=u6M6ECGok1o
 +
 +
- SUS
 
==See Also==
 
==See Also==
 +
 
{{AMC8 box|year=2017|num-b=20|num-a=22}}
 
{{AMC8 box|year=2017|num-b=20|num-a=22}}
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−
{{MAA Notice}}
 

Latest revision as of 15:11, 2 March 2024

Problem

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

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

Solution

There are $2$ cases to consider:

Case $1$: $2$ of $a$, $b$, and $c$ are positive and the other is negative. Without loss of generality (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, we can assume that $a$ and $b$ are negative and $c$ is positive. In this case, we have that \[\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}=-1-1+1+1=0.\]

Note these are the only valid cases, for neither $3$ negatives nor $3$ positives would work as they cannot sum up to $0$. In both cases, we get that the given expression equals $\boxed{\textbf{(A)}\ 0}$.

Video Solution (CREATIVE THINKING + ANALYSIS!!!)

https://youtu.be/wPVQwzA-2hU

~Education, the Study of Everything

Video Solution by OmegaLearn

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

Video Solutions

https://youtu.be/FUEHirfk-tw

https://youtu.be/V9wCBTwvIZo

- Happytwin

https://youtu.be/xN0dnJC1hv8

~savannahsolver

https://www.youtube.com/watch?v=u6M6ECGok1o

- SUS

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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