Difference between revisions of "2016 AMC 10A Problems/Problem 4"
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The value, by definition, is <cmath>\frac{3}{8}-\left(-\frac{2}{5}\right)\lfloor{\frac{3}{8}*\frac{-5}{2}\rfloor=-\frac{15}{16}}=\frac{3}{8}-\frac{2}{5}=\boxed{\textbf{(B) } -\frac{1}{40}.}</cmath> | The value, by definition, is <cmath>\frac{3}{8}-\left(-\frac{2}{5}\right)\lfloor{\frac{3}{8}*\frac{-5}{2}\rfloor=-\frac{15}{16}}=\frac{3}{8}-\frac{2}{5}=\boxed{\textbf{(B) } -\frac{1}{40}.}</cmath> | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC10 box|year=2016|ab=A|num-b=2|num-a=4}} | ||
| + | {{MAA Notice}} | ||
Revision as of 18:30, 3 February 2016
Problem
The remainder can be defined for all real numbers 
and 
with 
by ![\[\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor\]](http://latex.artofproblemsolving.com/9/5/c/95c21b1ce04c5f3c86f558991d39aa35a13bca21.png)
where 
denotes the greatest integer less than or equal to 
. What is the value of 
?
Solution
The value, by definition, is ![\[\frac{3}{8}-\left(-\frac{2}{5}\right)\lfloor{\frac{3}{8}*\frac{-5}{2}\rfloor=-\frac{15}{16}}=\frac{3}{8}-\frac{2}{5}=\boxed{\textbf{(B) } -\frac{1}{40}.}\]](http://latex.artofproblemsolving.com/d/2/c/d2ca06774f3be48b85c7660adbc5828a37071a9a.png)
See Also
| 2016 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
