Question

对所有实数 $x$ 和满足 $y \neq 0$ 的实数 $y$ ，余数定义为其中 $\left \lfloor \frac{x}{y} \right \rfloor$ 表示小于或等于 $\frac{x}{y}$ 的最大整数。求 $\text{rem} \left(\frac{3}{8}, -\frac{2}{5} \right)$ 的值。 $\operatorname{rem}(x,y)=x-y\left\lfloor\frac{x}{y}\right\rfloor,$

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

$-\dfrac{3}{8}$

$-\dfrac{1}{40}$

$0$

$\dfrac{3}{8}$

$\dfrac{31}{40}$

答案：B

知识点：取整函数分数

难度评级：1140

解答：

所以正确答案是 B。 $\begin{aligned} \text{rem} \left(\dfrac{3}{8}, -\dfrac{2}{5}\right) &= \dfrac{3}{8} \\ &{}+ \dfrac{2}{5} \left \lfloor \dfrac{\frac{3}{8}}{-\frac{2}{5}}\right \rfloor \\ &= \dfrac{3}{8} \\ &{}+ \dfrac{2}{5} \left \lfloor -\dfrac{15}{16} \right \rfloor \\ &= \dfrac{3}{8} + \dfrac{2}{5} \cdot -1 \\ &= \dfrac{3}{8} - \dfrac{2}{5} \\ &= - \dfrac{1}{40} \end{aligned}$

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 .

2016 AMC 10A 第 4 题

答案：B

难度评级：1140

解答：

其他年份的第 4 题