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