2016 AMC 10A Problem 4

First compute the quotient:
$$\frac{x}{y} = \frac{3/8}{-2/5} = \frac{3}{8} \cdot \left(-\frac{5}{2}\right) = -\frac{15}{16}.$$

The floor of $-\tfrac{15}{16}$ is the greatest integer that is at most $-0.9375$, which is $-1$ (not $0$, since $0 > -0.9375$).

Now apply the definition:
$$\text{rem}\left(\tfrac{3}{8}, -\tfrac{2}{5}\right) = \frac{3}{8} - \left(-\frac{2}{5}\right)(-1) = \frac{3}{8} - \frac{2}{5} = \frac{15}{40} - \frac{16}{40} = -\frac{1}{40}.$$

The answer is $\boxed{\textbf{(B)}\ -\frac{1}{40}}$.

The problem is a definition-following exercise whose only danger is the floor of a negative number: floor always rounds down on the number line, so $\lfloor -0.9 \rfloor = -1$. A sanity check: this remainder convention always produces a result with the same sign as $y$ and smaller in magnitude, and $-\tfrac{1}{40}$ fits.

• Taking the floor of -15/16 to be 0 (rounding toward zero) and answering 3/8.
• Sign slips when substituting: $-y \cdot \lfloor x/y \rfloor = -(-\tfrac25)(-1) = -\tfrac25$, not $+\tfrac25$ (which gives the distractor $\tfrac{31}{40}$).