Difference between revisions of "2017 AMC 8 Problems/Problem 21"
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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. 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> | ||
− | 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== | ||
https://youtu.be/FUEHirfk-tw | https://youtu.be/FUEHirfk-tw | ||
− | https://youtu.be/V9wCBTwvIZo | + | |
+ | https://youtu.be/V9wCBTwvIZo - Happytwin | ||
+ | |||
https://youtu.be/7an5wU9Q5hk?t=2362 | https://youtu.be/7an5wU9Q5hk?t=2362 | ||
+ | |||
+ | https://youtu.be/xN0dnJC1hv8 | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 11:01, 10 March 2022
Contents
Problem
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
Case : of , , and are negative and the other is positive. Without loss of generality, 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
https://youtu.be/V9wCBTwvIZo - Happytwin
https://youtu.be/7an5wU9Q5hk?t=2362
~savannahsolver
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.