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

m (Solution 1)
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In both cases, we get that the given expression equals <math>\boxed{\textbf{(A)}\ 0}</math>.
 
In both cases, we get that the given expression equals <math>\boxed{\textbf{(A)}\ 0}</math>.
 
==Solution 2==
 
 
Assuming numbers:
 
 
WLOG <math>a=1, b=2,</math> and <math>c=-3</math> (Other numbers can apply for <math>a, b,</math> and <math>c</math> as long as their sum is <math>0</math>.) . Then plug <math>a, b,</math> and <math>c</math> into the given equation. The result is always <math>\boxed{\textbf{(A)}\ 0}</math>.
 
 
  
 
==See Also==
 
==See Also==

Revision as of 15:03, 4 November 2018

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

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
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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