Difference between revisions of "2003 Indonesia MO Problems/Problem 3"
Rockmanex3 (talk | contribs) (Solution to Problem 3 -- floors and ceilings) |
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− | For the positive case, if <math>x = \sqrt{a},</math> then the equation results in <math>2a = 2003.</math> Since the equation does not have an integral solution, <math>x \ne \sqrt{a}.</math> If we let <math>\sqrt{a} < x < \sqrt{a+1}.</math> That means <math>a + a + 1 = 2003,</math> and solving the equation yields <math>a = 1001.</math> | + | For the positive case, if <math>x = \sqrt{a},</math> then the equation results in <math>2a = 2003.</math> Since the equation does not have an integral solution, <math>x \ne \sqrt{a}.</math> If we let <math>\sqrt{a} < x < \sqrt{a+1}.</math> That means <math>a + a + 1 = 2003,</math> and solving the equation yields <math>a = 1001.</math> For confirmation, <math>1001 \le \lfloor x^2 \rfloor < 1002</math> and <math>1001 < \lceil x^2 \rceil \le 1002</math>, so <math>2002 < \lfloor x^2 \rfloor + \lceil x^2 \rceil < 2004</math>. Since <math>\lfloor x^2 \rfloor + \lceil x^2 \rceil</math> can only have integral values, <math>\lfloor x^2 \rfloor + \lceil x^2 \rceil = 2003</math>. |
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− | For the negative case, if <math>x = -\sqrt{a},</math> then the equation results in <math>2a = 2003.</math> This also does not have an integral solution, so <math>x \ne \sqrt{a}.</math> If we let <math>-\sqrt{a+1} < x < -\sqrt{a}.</math> That means <math>a+1 + a = 2003,</math> and this equation also yields <math>a = 1001.</math> | + | For the negative case, if <math>x = -\sqrt{a},</math> then the equation results in <math>2a = 2003.</math> This also does not have an integral solution, so <math>x \ne \sqrt{a}.</math> If we let <math>-\sqrt{a+1} < x < -\sqrt{a}.</math> That means <math>a+1 + a = 2003,</math> and this equation also yields <math>a = 1001.</math> For confirmation, <math>1001 < \lfloor x^2 \rfloor \le 1002</math> and <math>1001 \le \lceil x^2 \rceil < 1002</math>, so <math>2002 < \lfloor x^2 \rfloor + \lceil x^2 \rceil < 2004</math>. Since <math>\lfloor x^2 \rfloor + \lceil x^2 \rceil</math> can only have integral values, <math>\lfloor x^2 \rfloor + \lceil x^2 \rceil = 2003</math>. |
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Latest revision as of 10:50, 17 March 2020
Problem
Find all real solutions of the equation .
[Note: For any real number , is the largest integer less than or equal to , and denote the smallest integer more than or equal to .]
Solution
There are two cases to consider -- one where is positive and one where is negative.
For the positive case, if then the equation results in Since the equation does not have an integral solution, If we let That means and solving the equation yields For confirmation, and , so . Since can only have integral values, .
For the negative case, if then the equation results in This also does not have an integral solution, so If we let That means and this equation also yields For confirmation, and , so . Since can only have integral values, .
In interval notation, the solutions of the equation are
See Also
2003 Indonesia MO (Problems) | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 | Followed by Problem 4 |
All Indonesia MO Problems and Solutions |