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

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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>.

Revision as of 12:27, 15 December 2016

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

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


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 $\boxed{\textbf{(E)}\ \text{for no non-zero real numbers } x}$.

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