Difference between revisions of "1986 AHSME Problems/Problem 7"

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==Solution==
 
==Solution==
If <math>x \leq 2</math>, then <math>\floor{x} + \ceil{x} \leq 2+2 \lt 5</math>, so there are no solutions with <math>x \leq 2</math>. If <math>x \geq 3</math>, then <math>\floor{x} + \ceil{x} \geq 3+3</math>, so there are also no solutions here. Finally, if <math>2 \lt x \lt 3</math>, then <math>\floor{x} + \ceil{x} = 2 + 3 = 5</math>, so the solution set is <math>\boxed{E}</math>.
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If <math>x \leq 2</math>, then <math>\lfloor x \rfloor + \lceil x \rceil \leq 2+2 < 5</math>, so there are no solutions with <math>x \leq 2</math>. If <math>x \geq 3</math>, then <math>\lfloor x \rfloor + \lceil x \rceil \geq 3+3</math>, so there are also no solutions here. Finally, if <math>2<x<3</math>, then <math>\lfloor x \rfloor + \lceil x \rceil = 2 + 3 = 5</math>, so the solution set is <math>\boxed{E}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 18:20, 1 April 2018

Problem

The sum of the greatest integer less than or equal to $x$ and the least integer greater than or equal to $x$ is $5$. The solution set for $x$ is

$\textbf{(A)}\ \Big\{\frac{5}{2}\Big\}\qquad \textbf{(B)}\ \big\{x\ |\ 2 \le x \le 3\big\}\qquad \textbf{(C)}\ \big\{x\ |\ 2\le x < 3\big\}\qquad\\  \textbf{(D)}\ \Big\{x\ |\ 2 < x\le 3\Big\}\qquad \textbf{(E)}\ \Big\{x\ |\ 2 < x < 3\Big\}$


Solution

If $x \leq 2$, then $\lfloor x \rfloor + \lceil x \rceil \leq 2+2 < 5$, so there are no solutions with $x \leq 2$. If $x \geq 3$, then $\lfloor x \rfloor + \lceil x \rceil \geq 3+3$, so there are also no solutions here. Finally, if $2<x<3$, then $\lfloor x \rfloor + \lceil x \rceil = 2 + 3 = 5$, so the solution set is $\boxed{E}$.

See also

1986 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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