Difference between revisions of "2016 AMC 12A Problems/Problem 3"
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<cmath>=\frac{3}{8}-\frac{2}{5}</cmath> | <cmath>=\frac{3}{8}-\frac{2}{5}</cmath> | ||
<cmath>=\boxed{\textbf{(B)}-\frac{1}{40}}</cmath> | <cmath>=\boxed{\textbf{(B)}-\frac{1}{40}}</cmath> | ||
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| + | ==See Also== | ||
| + | {{AMC12 box|year=2016|ab=A|num-b=2|num-a=4}} | ||
| + | {{MAA Notice}} | ||
Revision as of 23:46, 3 February 2016
Problem
The remainder function can be defined for all real numbers 
and 
with 
by
,
where 
denotes the greatest integer less than or equal to 
. What is the value of 
?
Solution
![\[\text{rem}\left(\frac{3}{8},-\frac{2}{5}\right)\]](http://latex.artofproblemsolving.com/4/b/c/4bc95cf8469a4cfdb6d3b62fdf68c7f9a237ba3b.png)
![\[=\frac{3}{8}-\left(-\frac{2}{5}\right)\Big\lfloor\frac{\frac{3}{8}}{-\frac{2}{5}}\Big\rfloor\]](http://latex.artofproblemsolving.com/7/1/9/7199ee4077d312c29bb69bff4eb734c8d19afbee.png)
![\[=\frac{3}{8}+\left(\frac{2}{5}\right)\Big\lfloor -\frac{15}{16}\Big\rfloor\]](http://latex.artofproblemsolving.com/5/f/8/5f8b7c703033544a74733c002937ae58f2e4eeaf.png)
![\[=\frac{3}{8}+\left(\frac{2}{5}\right)\left(-1\right)\]](http://latex.artofproblemsolving.com/c/e/1/ce191ca7c81c2c3d36c3c92798070445a4a58ca3.png)
![\[=\frac{3}{8}-\frac{2}{5}\]](http://latex.artofproblemsolving.com/0/e/8/0e82baa90861d265d469d327e817acadd6168a85.png)
![\[=\boxed{\textbf{(B)}-\frac{1}{40}}\]](http://latex.artofproblemsolving.com/f/f/e/ffe2006d31961de958874647aa4dcb4b583657fd.png)
See Also
| 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. 
