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

Latest revision as of 11:13, 20 January 2018

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

$\begin{align} x+\lfloor y \rfloor +\{z\}=14.2,\\ \lfloor x \rfloor+\{y\} +z=15.3,\\ \{x\}+y +\lfloor z \rfloor=16.1. \end{align}$

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 $10(x+y+z)=5\cdot 45.6=\boxed{228}$ . View https://artofproblemsolving.com/texer/boixabdc , for a complete solution. We finally get x=6.5, y=7.6, z=8.7 . I'll solve a more general problem, see my blog https://artofproblemsolving.com/community/c573365.

@@ Line 2: / Line 2: @@
 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> given that
-<center><math>\begin{align}
+<center><cmath>\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}</math></center>
+\end{align}</cmath></center>
 ==Solution==
@@ Line 14: / Line 14: @@
 And thus <math>10(x+y+z)=5\cdot 45.6=\boxed{228}</math>.
+View https://artofproblemsolving.com/texer/boixabdc , for a complete solution. We finally get x=6.5, y=7.6, z=8.7 .
+I'll solve a more general problem, see my blog https://artofproblemsolving.com/community/c573365.
 ==See also==
+{{ARML box|year=2006|state=Alabama|num-b=5|num-a=7}}