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| − | | + | #REDIRECT [[2016 AMC 10A Problems/Problem 4]] |
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| − | ==Problem==
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| − | The remainder function can be defined for all real numbers <math>x</math> and <math>y</math> with <math>y\ne 0</math> by
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| − | <math>\text{rem}(x,y)=x-y\Big\lfloor\frac{x}{y}\Big\rfloor</math>,
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| − | where <math>\Big\lfloor\frac{x}{y}\Big\rfloor</math> denotes the greatest integer less than or equal to <math>\frac{x}{y}</math>. What is the value of <math>\text{rem}\left(\frac{3}{8},-\frac{2}{5}\right)</math>?
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| − | <math>\textbf{(A)}\ -\frac{3}{8}\qquad\textbf{(B)}\ -\frac{1}{40}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ \frac{3}{8}\qquad\textbf{(E)}\ \frac{31}{40}</math>
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| − | ==Solution==
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| − | <cmath>\text{rem}\left(\frac{3}{8},-\frac{2}{5}\right)</cmath>
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| − | <cmath>=\frac{3}{8}-\left(-\frac{2}{5}\right)\Big\lfloor\frac{\frac{3}{8}}{-\frac{2}{5}}\Big\rfloor</cmath>
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| − | <cmath>=\frac{3}{8}+\left(\frac{2}{5}\right)\Big\lfloor -\frac{15}{16}\Big\rfloor</cmath>
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| − | <cmath>=\frac{3}{8}+\left(\frac{2}{5}\right)\left(-1\right)</cmath>
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| − | <cmath>=\frac{3}{8}-\frac{2}{5}</cmath>
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| − | <cmath>=\boxed{\textbf{(B)}-\frac{1}{40}}</cmath>
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| − | ==See Also==
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| − | {{AMC12 box|year=2016|ab=A|num-b=2|num-a=4}}
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| − | {{MAA Notice}}
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