AMC12 2016 A

AMC12 2016 A · Q3

AMC12 2016 A · Q3. It mainly tests Inequalities with floors/ceilings (basic), Fractions.

The remainder function can be defined for all real numbers $x$ and $y$ with $y\ne 0$ by $\mathrm{rem}(x,y)=x-y\left\lfloor \frac{x}{y}\right\rfloor,$ 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 $\mathrm{rem}\!\left(\frac{3}{8},-\frac{2}{5}\right)$?

余数函数对所有实数 $x$ 和 $y$（其中 $y\ne 0$）定义为 $\mathrm{rem}(x,y)=x-y\left\lfloor \frac{x}{y}\right\rfloor,$ 其中 $\left\lfloor \frac{x}{y}\right\rfloor$ 表示不大于 $\frac{x}{y}$ 的最大整数。求 $\mathrm{rem}\!\left(\frac{3}{8},-\frac{2}{5}\right)$ 的值。

(A) $-\frac{3}{8}$ $-\frac{3}{8}$

(B) $-\frac{1}{40}$ $-\frac{1}{40}$

(D) $\frac{3}{8}$ $\frac{3}{8}$

(E) $\frac{31}{40}$ $\frac{31}{40}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): $\frac{3}{8}-\left(-\frac{2}{5}\right)\left[\frac{\frac{3}{8}}{-\frac{2}{5}}\right]=\frac{3}{8}+\frac{2}{5}\left[\frac{15}{-16}\right]=\frac{3}{8}+\frac{2}{5}(-1)=-\frac{1}{40}$

答案（B）： $\frac{3}{8}-\left(-\frac{2}{5}\right)\left[\frac{\frac{3}{8}}{-\frac{2}{5}}\right]=\frac{3}{8}+\frac{2}{5}\left[\frac{15}{-16}\right]=\frac{3}{8}+\frac{2}{5}(-1)=-\frac{1}{40}$

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