Difference between revisions of "2017 AMC 8 Problems/Problem 21"
m (→Solution 1) |
Sharpwhiz17 (talk | contribs) m |
||
| (34 intermediate revisions by 20 users not shown) | |||
| Line 1: | Line 1: | ||
| − | ==Problem | + | ==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>? | ||
| Line 5: | Line 5: | ||
<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== |
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, 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. 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> | ||
| − | 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 | + | ==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 | ||
| + | |||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | |||
| + | https://www.youtube.com/watch?v=xN0dnJC1hv8&list=PL9t_YdPye18Wk-urEfwl0kex_RFg1lxPW&index=23 | ||
| + | |||
| + | - Rayansh Mankad | ||
| + | |||
| + | ~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}} | ||
| − | |||
| − | |||
Latest revision as of 18:53, 3 January 2025
Contents
Problem
Suppose 
, 
, and 
are nonzero real numbers, and 
. What are the possible value(s) for 
?
Solution
There are 
cases to consider:
Case 
: of , , and are positive and the other is negative. Without loss of generality (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)
Note these are the only valid cases, for neither 
negatives nor positives would work as they cannot sum up to 
. In both cases, we get that the given expression equals 
.
Video Solution (CREATIVE THINKING + ANALYSIS!!!)
~Education, the Study of Everything
Video Solution by OmegaLearn
https://youtu.be/7an5wU9Q5hk?t=2362
Video Solutions
- Happytwin
Video Solution by WhyMath
https://www.youtube.com/watch?v=xN0dnJC1hv8&list=PL9t_YdPye18Wk-urEfwl0kex_RFg1lxPW&index=23
- Rayansh Mankad
~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 | ||
