Difference between revisions of "2017 AMC 8 Problems/Problem 21"
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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>\ | + | <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. | + | 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>. |
− | ==Video Solution== | + | ==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== |
− | + | https://youtu.be/wPVQwzA-2hU | |
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+ | ~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 | ||
+ | |||
+ | ~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}} | ||
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Latest revision as of 16:11, 2 March 2024
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
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
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
~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 |