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

m (Solution 2)
m (Video Solutions)
(26 intermediate revisions by 17 users not shown)
Line 1: Line 1:
==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>\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>\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>
  
 
==Solution 1==
 
==Solution 1==
Line 9: Line 9:
 
There are <math>2</math> cases to consider:
 
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 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. WLOG 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>
  
In both cases, we get that the given expression equals <math>\boxed{\textbf{(A)}\ 0}</math>.
+
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>.
  
==Solution 2==
+
==Video Solution by OmegaLearn==
 +
https://youtu.be/7an5wU9Q5hk?t=2362
  
Assuming numbers:
+
==Video Solutions==
 +
https://youtu.be/FUEHirfk-tw
  
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>.
+
https://youtu.be/V9wCBTwvIZo
 +
 
 +
- Happytwin
 +
 
 +
https://youtu.be/xN0dnJC1hv8
 +
 
 +
~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2017|num-b=20|num-a=22}}
 
{{AMC8 box|year=2017|num-b=20|num-a=22}}
  
 +
[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:46, 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

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

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