2016 AMC 10A Problems/Problem 4

Problem

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

$\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}$

Solution

The value, by definition, is $\begin{align*} \text{rem}\left(\frac{3}{8},-\frac{2}{5}\right) &= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{\frac{3}{8}}{-\frac{2}{5}}\right\rfloor \\ &= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{3}{8}\times\frac{-5}{2}\right\rfloor \\ &= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{-15}{16}\right\rfloor\\ &= \frac{3}{8}-\left(-\frac{2}{5}\right)\left(-1\right)\\ &= \frac{3}{8}-\frac{2}{5}\\ &= \boxed{\textbf{(B) } -\frac{1}{40}}. \end{align*}$

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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All AMC 10 Problems and Solutions

2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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 12 Problems and Solutions

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2016 AMC 10A Problems/Problem 4

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