Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 10, 2011"
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==Solution== | ==Solution== | ||
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+ | <math>5x = \lfloor x + 1 \rfloor + \lceil x + 2 \rceil</math>, <math>0 < x < 2</math>. | ||
+ | If <math>x</math> is an integer, <math>5x = x + 1 + x + 2 \implies x = 1</math>. | ||
+ | If <math>x</math> isn't an integer, <math>5x = \lfloor x + 1 \rfloor + \lceil x + 2 \rceil = \lfloor x \rfloor + 1 + \lfloor x \rfloor + 2 + 1 \implies 5x = 2 \lfloor x \rfloor + 4</math>. | ||
+ | Since <math>\lfloor x \rfloor = 0</math> or <math>1</math>, <math>5x = 4 \implies x = \dfrac{4}{5}</math> or <math>5x = 6 \implies x = \dfrac{6}{5}</math>. Checking all our solutions, they all work. Thus the solutions are <math>x = \dfrac{4}{5}</math>, <math>1</math>, or <math>\dfrac{6}{5}</math>, and the sum of them all is <math>\boxed{3}</math>. |
Latest revision as of 22:25, 9 June 2011
Problem
AoPSWiki:Problem of the Day/June 10, 2011
Solution
, . If is an integer, . If isn't an integer, . Since or , or . Checking all our solutions, they all work. Thus the solutions are , , or , and the sum of them all is .