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2016 AMC 12A Problem 3

Problem 3 of 25EasierAlgebra

The remainder function can be defined for all real numbers xx and yy with y≠0y\neq 0 by rem(x,y)=x−y⌊xy⌋, \text{rem}(x,y)=x-y\left\lfloor \dfrac{x}{y}\right\rfloor, where ⌊xy⌋\left\lfloor \dfrac{x}{y}\right\rfloor denotes the greatest integer less than or equal to xy.\dfrac{x}{y}. What is the value of rem(38,−25)?\text{rem}\left(\dfrac{3}{8},-\dfrac{2}{5}\right)?

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Solution

First, xy=38−25=38⋅(−52)=−1516, \begin{gathered} \dfrac{x}{y}=\dfrac{\frac{3}{8}}{-\frac{2}{5}}\\ =\dfrac{3}{8}\cdot\left(-\dfrac{5}{2}\right)\\ =-\dfrac{15}{16}, \end{gathered} and ⌊−1516⌋=−1.\left\lfloor -\dfrac{15}{16}\right\rfloor=-1. Therefore rem(38,−25)=38−(−25)(−1)=38−25=15−1640=−140. \begin{gathered} \text{rem}\left(\dfrac{3}{8},-\dfrac{2}{5}\right)\\ =\dfrac{3}{8}-\left(-\dfrac{2}{5}\right)(-1)\\ =\dfrac{3}{8}-\dfrac{2}{5}\\ =\dfrac{15-16}{40}\\ =-\dfrac{1}{40}. \end{gathered} Thus, the correct answer is B.
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