Difference between revisions of "1975 AHSME Problems/Problem 7"
m (→Solution) |
Smallstone9 (talk | contribs) m (→Problem) |
||
| (One intermediate revision by one other user not shown) | |||
| Line 1: | Line 1: | ||
| − | For which non-zero real numbers <math>x</math> is <math>\frac{|x-|x|\-|}{x}</math> a positive | + | ==Problem== |
| + | |||
| + | 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 \\ | ||
| Line 5: | Line 7: | ||
\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 \\ | ||
| − | \textbf{(E)}\ \text{for no non-zero real numbers } x </math> | + | \textbf{(E)}\ \text{for no non-zero real numbers } x </math> |
| − | |||
==Solution== | ==Solution== | ||
| Line 13: | Line 14: | ||
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>. | 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>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME box|year=1975|num-b=6|num-a=8}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 22:43, 6 October 2021
Problem
For which non-zero real numbers 
is 
a positive integer?
Solution
Solution by e_power_pi_times_i
Notice that if is negative, then the whole thing would amount to a negative number. Also notice that if is positive, then 
would be 
, hence the whole thing would amount to . Therefore, is positive 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
