Difference between revisions of "2017 AMC 8 Problems/Problem 21"
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<math>\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</math> | <math>\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</math> | ||
| − | ==Solution== | + | ==Solution 1== |
There are <math>2</math> cases to consider: | There are <math>2</math> cases to consider: | ||
Revision as of 17:48, 30 December 2018
Problem 21
Suppose 
, 
, and 
are nonzero real numbers, and 
. What are the possible value(s) for 
?
Solution 1
There are 
cases to consider:
Case 
: of , , and are positive and the other is negative. WLOG, we can assume that and are positive and is negative. In this case, we have that ![\[\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}=1+1-1-1=0.\]](http://latex.artofproblemsolving.com/1/1/2/1129f25a010f4ba94c660c13c8035a26721937b2.png)
Case : of , , and are negative and the other is positive. WLOG, we can assume that and are negative and is positive. In this case, we have that ![\[\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}=-1-1+1+1=0.\]](http://latex.artofproblemsolving.com/6/b/3/6b3f2333ac9e5b385d2d7c7342575cc5daa8b2ce.png)
In both cases, we get that the given expression equals 
.
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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