2016 AMC 10A Problema 4

Question

La función residuo se puede definir para todos los números reales

x

y

y

con

y \neq 0

mediante

\operatorname{rem}(x,y)=x-y\left\lfloor\frac{x}{y}\right\rfloor,

donde

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

denota el mayor entero menor o igual que

\frac{x}{y}

. ¿Cuál es el valor de

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

?

The remainder function can be defined for all real numbers

x

and

y

with

y \neq 0

by

\operatorname{rem}(x,y)=x-y\left\lfloor\frac{x}{y}\right\rfloor,

where

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

denotes the greatest integer less than or equal to

\frac{x}{y}.

What is the value of

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

-\dfrac{3}{8}

-\dfrac{1}{40}

0

\dfrac{3}{8}

\dfrac{31}{40}

Calcula

\frac{x}{y}

antes de aplicar la función piso.

Compute

\frac{x}{y}

before applying the floor function

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

.

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

Usando la fórmula, obtenemos

\begin{aligned} \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{aligned}

Por lo tanto, la respuesta correcta es B.

Using the formula, we get

\begin{aligned} \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{aligned}

Thus, the correct answer is B .

El Problema 4 en otros años