Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 10, 2011"
(new problem talk page) |
|||
Line 2: | Line 2: | ||
{{:AoPSWiki:Problem of the Day/June 10, 2011}} | {{:AoPSWiki:Problem of the Day/June 10, 2011}} | ||
==Solution== | ==Solution== | ||
− | {{ | + | |
+ | <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
.