Difference between revisions of "1977 AHSME Problems/Problem 8"
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<math>\dfrac{x}{|x|} = 1</math> or <math>-1</math> depending whether <math>x</math> is positive or negative. If <math>a</math>, <math>b</math>, and <math>c</math> are positive, then the entire thing amounts to <math>4</math>. If one of the three is negative and the other two positive, the answer is <math>0</math>. If two of the three is negative and one is positive, the answer is <math>0</math>. If all three are negative, the answer is <math>-4</math>. Therefore the set is <math>\boxed{\textbf{(B)}\ \{-4,0,4\}}</math>. | <math>\dfrac{x}{|x|} = 1</math> or <math>-1</math> depending whether <math>x</math> is positive or negative. If <math>a</math>, <math>b</math>, and <math>c</math> are positive, then the entire thing amounts to <math>4</math>. If one of the three is negative and the other two positive, the answer is <math>0</math>. If two of the three is negative and one is positive, the answer is <math>0</math>. If all three are negative, the answer is <math>-4</math>. Therefore the set is <math>\boxed{\textbf{(B)}\ \{-4,0,4\}}</math>. | ||
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+ | This is very similar to https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_21 |
Latest revision as of 00:43, 3 October 2020
Problem 8
For every triple of non-zero real numbers, form the number
.
The set of all numbers formed is
Solution
Solution by e_power_pi_times_i
or
depending whether
is positive or negative. If
,
, and
are positive, then the entire thing amounts to
. If one of the three is negative and the other two positive, the answer is
. If two of the three is negative and one is positive, the answer is
. If all three are negative, the answer is
. Therefore the set is
.
This is very similar to https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_21