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Difference between revisions of "1975 AHSME Problems/Problem 7"

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For which non-zero real numbers <math>x</math> is <math>\frac{|x-|x|\-|}{x}</math> a positive integers?  
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==Problem==
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For which non-zero real numbers <math>x</math> is <math>\frac{|x-|x|\-|}{x}</math> a positive integer?  
  
 
<math>\textbf{(A)}\ \text{for negative } x \text{ only} \qquad \\
 
<math>\textbf{(A)}\ \text{for negative } x \text{ only} \qquad \\
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\textbf{(C)}\ \text{only for } x \text{ an even integer} \qquad \\
 
\textbf{(C)}\ \text{only for } x \text{ an even integer} \qquad \\
 
\textbf{(D)}\ \text{for all non-zero real numbers } x \\
 
\textbf{(D)}\ \text{for all non-zero real numbers } x \\
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\textbf{(E)}\ \text{for no non-zero real numbers } x  </math>
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\textbf{(E)}\ \text{for no non-zero real numbers } x  </math>
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==Solution==
 
==Solution==
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Notice that if <math>x</math> is negative, then the whole thing would amount to a negative number. Also notice that if <math>x</math> is positive, then <math>|x-|x|\-|</math> would be <math>0</math>, hence the whole thing would amount to <math>0</math>. Therefore, <math>\frac{|x-|x|\-|}{x}</math> is positive <math>\boxed{\textbf{(E) } \text{for no non-zero real numbers}\ \x}</math>.
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Notice that if <math>x</math> is negative, then the whole thing would amount to a negative number. Also notice that if <math>x</math> is positive, then <math>|x-|x|\-|</math> would be <math>0</math>, hence the whole thing would amount to <math>0</math>. Therefore, <math>\frac{|x-|x|\-|}{x}</math> is positive <math>\boxed{\textbf{(E)}\ \text{for no non-zero real numbers } x}</math>.
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==See Also==
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{{AHSME box|year=1975|num-b=6|num-a=8}}
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{{MAA Notice}}

Latest revision as of 22:43, 6 October 2021

Problem

For which non-zero real numbers $x$ is $\frac{|x-|x|\-|}{x}$ a positive integer?

$\textbf{(A)}\ \text{for negative } x \text{ only} \qquad \\ \textbf{(B)}\ \text{for positive } x \text{ only} \qquad \\ \textbf{(C)}\ \text{only for } x \text{ an even integer} \qquad \\ \textbf{(D)}\ \text{for all non-zero real numbers } x \\ \textbf{(E)}\ \text{for no non-zero real numbers } x$

Solution

Solution by e_power_pi_times_i


Notice that if $x$ is negative, then the whole thing would amount to a negative number. Also notice that if $x$ is positive, then $|x-|x|\-|$ would be $0$, hence the whole thing would amount to $0$. Therefore, $\frac{|x-|x|\-|}{x}$ is positive $\boxed{\textbf{(E)}\ \text{for no non-zero real numbers } x}$.

See Also

1975 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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