Difference between revisions of "1986 AHSME Problems/Problem 7"
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==Solution== | ==Solution== | ||
| − | If <math>x \leq 2</math>, then <math>\ | + | 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 17:20, 1 April 2018
Problem
The sum of the greatest integer less than or equal to 
and the least integer greater than or equal to is 
. The solution set for is
Solution
If 
, then 
, so there are no solutions with . If 
, then 
, so there are also no solutions here. Finally, if 
, then 
, so the solution set is 
.
See also
| 1986 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. 
