Difference between revisions of "2006 Alabama ARML TST Problems/Problem 6"

Revision as of 07:21, 15 April 2008

Let $\lfloor a \rfloor$ be the greatest integer less than or equal to $a$ and let $\{a\}=a-\lfloor a \rfloor$ . Find $10(x+y+z)$ given that

$\begin{align}

x+\lfloor y \rfloor +\{z\}=14.2,\\ \lfloor x \rfloor+\{y\} +z=15.3,\\ \{x\}+y +\lfloor z \rfloor=16.1.

\end{align}$ (Error compiling LaTeX. Unknown error_msg)

Let's add all three equations:

$x+\lfloor y \rfloor +\{z\}+\lfloor x \rfloor+\{y\} +z+\{x\}+y +\lfloor z \rfloor=x+x+y+y+z+z=2(x+y+z)=45.6$

And thus $10(x+y+z)=5\cdot 45.6=\boxed{228}$ .

Revision as of 07:20, 15 April 2008 (view source) 1=2 (talk \| contribs) (New page: ==Problem== Let <math>\lfloor a \rfloor</math> be the greatest integer less than or equal to <math>a</math> and let <math>\{a\}=a-\lfloor a \rfloor</math>. Find <math>10(x+y+z)</math> give...)		Revision as of 07:21, 15 April 2008 (view source) 1=2 (talk \| contribs) (→‎See also) Newer edit →
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	==See also==		==See also==
		+	{{ARML box\|year=2006\|state=Alabama\|num-b=5\|num-a=7}}