2006 Alabama ARML TST Problems/Problem 6


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.

See also

2006 Alabama ARML TST (Problems)
Preceded by:
Problem 5
Followed by:
Problem 7
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