Difference between revisions of "2016 AMC 10A Problems/Problem 4"
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The value, by definition, is <cmath>\begin{align*} | The value, by definition, is <cmath>\begin{align*} | ||
− | \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{3}{8}\times\frac{-5}{2}\right\rfloor | + | \text{rem}\left(\frac{3}{8},-\frac{2}{5}\right) |
+ | &= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{\frac{3}{8}}{-\frac{2}{5}}\right\rfloor \\ | ||
+ | &= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{3}{8}\times\frac{-5}{2}\right\rfloor \\ | ||
&= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{-15}{16}\right\rfloor\\ | &= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{-15}{16}\right\rfloor\\ | ||
&= \frac{3}{8}-\left(-\frac{2}{5}\right)\left(-1\right)\\ | &= \frac{3}{8}-\left(-\frac{2}{5}\right)\left(-1\right)\\ |
Revision as of 11:50, 4 February 2016
Problem
The remainder 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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.