2016 AMC 12A Problema 3

Question

La función resto puede definirse para todos los números reales

x

e

y

con

y\neq 0

mediante

\text{rem}(x,y)=x-y\left\lfloor \dfrac{x}{y}\right\rfloor,

donde

\left\lfloor \dfrac{x}{y}\right\rfloor

denota el mayor entero menor o igual que

\dfrac{x}{y}.

¿Cuál es el valor de

\text{rem}\left(\dfrac{3}{8},-\dfrac{2}{5}\right)

?

The remainder function can be defined for all real numbers

x

and

y

with

y\neq 0

by

\text{rem}(x,y)=x-y\left\lfloor \dfrac{x}{y}\right\rfloor,

where

\left\lfloor \dfrac{x}{y}\right\rfloor

denotes the greatest integer less than or equal to

\dfrac{x}{y}.

What is the value of

\text{rem}\left(\dfrac{3}{8},-\dfrac{2}{5}\right)?

-\dfrac{3}{8}

-\dfrac{1}{40}

0

\dfrac{3}{8}

\dfrac{31}{40}

Calcula primero

\dfrac{x}{y}=\dfrac{\frac{3}{8}}{-\frac{2}{5}}

.

Compute

\dfrac{x}{y}=\dfrac{\frac{3}{8}}{-\frac{2}{5}}

first

\left\lfloor -\dfrac{15}{16}\right\rfloor=-1

.

\left\lfloor -\dfrac{15}{16}\right\rfloor=-1

Primero,

\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}

y

\left\lfloor -\dfrac{15}{16}\right\rfloor=-1.

Por consiguiente

\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}

Por lo tanto, la respuesta correcta es B.

First,

\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

\left\lfloor -\dfrac{15}{16}\right\rfloor=-1.

Therefore

\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.

El Problema 3 en otros años