2016 AMC 10A Problem 4

Below is the professionally curated solution for Problem 4 of the 2016 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AMC 10A solutions, or check the answer key.

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Concepts:floor and ceiling functionsfraction

Difficulty rating: 1140

4.

The remainder can be defined for all real numbers xx and yy with y0y \neq 0 by rem(x,y)=xyxy\text{rem} (x ,y)=x-y\left \lfloor \dfrac{x}{y} \right \rfloorwhere xy\left \lfloor \frac{x}{y} \right \rfloor denotes the greatest integer less than or equal to xy.\frac{x}{y}. What is the value of rem(38,25)?\text{rem} \left(\frac{3}{8}, -\frac{2}{5} \right)?

38-\dfrac{3}{8}

140-\dfrac{1}{40}

00

38\dfrac{3}{8}

3140\dfrac{31}{40}

Solution:

Using the formula, we get rem(38,25)=38+253825=38+251516=38+251=3825=140 \begin{align*} \text{rem} \left(\dfrac{3}{8}, -\dfrac{2}{5}\right) &= \dfrac{3}{8} + \dfrac{2}{5} \left \lfloor \dfrac{\frac{3}{8}}{-\frac{2}{5}}\right \rfloor \\ &= \dfrac{3}{8} + \dfrac{2}{5} \left \lfloor -\dfrac{15}{16} \right \rfloor \\ &= \dfrac{3}{8} + \dfrac{2}{5} \cdot -1 \\ &= \dfrac{3}{8} - \dfrac{2}{5} \\ &= - \dfrac{1}{40} \end{align*} Thus, the correct answer is B .

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