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Difference between revisions of "1977 AHSME Problems/Problem 8"

(Created page with "==Solution== Solution by e_power_pi_times_i <math>\dfrac{x}{|x} = 1</math> or <math>-1</math> depending whether <math>x</math> is positive or negative. If <math>a</math>, <ma...")
 
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== Problem 8 ==
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For every triple <math>(a,b,c)</math> of non-zero real numbers, form the number <math>\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}</math>.
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The set of all numbers formed is
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<math>\textbf{(A)}\ {0} \qquad
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\textbf{(B)}\ \{-4,0,4\} \qquad
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\textbf{(C)}\ \{-4,-2,0,2,4\} \qquad
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\textbf{(D)}\ \{-4,-2,2,4\}\qquad
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\textbf{(E)}\ \text{none of these}  </math> 
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==Solution==
 
==Solution==
 
Solution by e_power_pi_times_i
 
Solution by e_power_pi_times_i
  
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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>.
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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>.
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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 23:43, 2 October 2020

Problem 8

For every triple $(a,b,c)$ of non-zero real numbers, form the number $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}$. The set of all numbers formed is

$\textbf{(A)}\ {0} \qquad \textbf{(B)}\ \{-4,0,4\} \qquad \textbf{(C)}\ \{-4,-2,0,2,4\} \qquad \textbf{(D)}\ \{-4,-2,2,4\}\qquad \textbf{(E)}\ \text{none of these}$


Solution

Solution by e_power_pi_times_i

$\dfrac{x}{|x|} = 1$ or $-1$ depending whether $x$ is positive or negative. If $a$, $b$, and $c$ are positive, then the entire thing amounts to $4$. If one of the three is negative and the other two positive, the answer is $0$. If two of the three is negative and one is positive, the answer is $0$. If all three are negative, the answer is $-4$. Therefore the set is $\boxed{\textbf{(B)}\ \{-4,0,4\}}$.

This is very similar to https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_21

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