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

Revision as of 19:44, 13 January 2023

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. 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 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

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

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

@@ Line 9: / Line 9: @@
 There are <math>2</math> cases to consider:
-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>
+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. Without loss of generality, 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>
 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>.

Page

Toolbox

Search