Difference between revisions of "2006 Alabama ARML TST Problems/Problem 6"
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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 | 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>< | + | <center><cmath>\begin{align} |
x+\lfloor y \rfloor +\{z\}=14.2,\\ | x+\lfloor y \rfloor +\{z\}=14.2,\\ | ||
\lfloor x \rfloor+\{y\} +z=15.3,\\ | \lfloor x \rfloor+\{y\} +z=15.3,\\ | ||
\{x\}+y +\lfloor z \rfloor=16.1. | \{x\}+y +\lfloor z \rfloor=16.1. | ||
| − | \end{align}</ | + | \end{align}</cmath></center> |
==Solution== | ==Solution== | ||
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And thus <math>10(x+y+z)=5\cdot 45.6=\boxed{228}</math>. | 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== | ==See also== | ||
{{ARML box|year=2006|state=Alabama|num-b=5|num-a=7}} | {{ARML box|year=2006|state=Alabama|num-b=5|num-a=7}} | ||
Latest revision as of 11:13, 20 January 2018
Problem
Let 
be the greatest integer less than or equal to 
and let 
. Find 
given that
Solution
Let's add all three equations:
And thus 
.
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 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
