Difference between revisions of "2016 AMC 10A Problems/Problem 4"
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&= \frac{3}{8}-\left(-\frac{2}{5}\right)\left(-1\right)\\ | &= \frac{3}{8}-\left(-\frac{2}{5}\right)\left(-1\right)\\ | ||
&= \frac{3}{8}-\frac{2}{5}\\ | &= \frac{3}{8}-\frac{2}{5}\\ | ||
| − | &= \boxed{\textbf{(B) } -\frac{1}{40}} | + | &= \boxed{\textbf{(B) } -\frac{1}{40}}. |
| − | \end{align*} | + | \end{align*}</cmath> |
==See Also== | ==See Also== | ||
Revision as of 16:43, 17 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 
See Also
| 2016 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 | ||
| 2016 AMC 12A (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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
