|
|
| (6 intermediate revisions by one other user not shown) |
| Line 1: |
Line 1: |
| − | ==Solution==
| + | #REDIRECT [[2016 AMC 10A Problems/Problem 4]] |
| − | <cmath>\text{rem}\left(\frac{3}{8},-\frac{2}{5}\right)</cmath>
| |
| − | <cmath>=\frac{3}{8}-\left(-\frac{2}{5}\right)\lfloor\frac{\frac{3}{8}}{-\frac{-2}{5}}\rfloor</cmath>
| |
| − | <cmath>=\frac{3}{8}+\left(\frac{2}{5}\right)\lfloor -\frac{15}{16}\rfloor</cmath>
| |
| − | <cmath>=\frac{3}{8}+\left(\frac{2}{5}\right)\left(-1\right)</cmath>
| |
| − | <cmath>=\frac{3}{8}-\frac{2}{5}</cmath>
| |
| − | <cmath>=\boxed{\textbf{(B)}-\frac{1}{40}}</cmath>
| |
| − | Notice that it would not matter if either <math>-\frac{2}{5}</math> or <math>\frac{2}{5}</math> were used.
| |
