2016 AMC 10A Problem 4

This is a "follow the definition" problem. The only danger is rounding a negative number down.

1. Here $x = \tfrac38$ and $y = -\tfrac25$. First find $\tfrac{x}{y}$. Dividing by a fraction means multiplying by its flip:
$$\frac{3}{8} \div \left(-\frac{2}{5}\right) = \frac{3}{8} \cdot \left(-\frac{5}{2}\right) = -\frac{15}{16}.$$
2. Now take the greatest integer less than or equal to $-\tfrac{15}{16}$. As a decimal, $-\tfrac{15}{16} = -0.9375$. On the number line it sits between $-1$ and $0$. The integer $0$ is bigger than $-0.9375$, so it is not allowed. The greatest integer that is not bigger is $-1$.
3. Put the pieces into the formula:
$$\text{rem} = x - y \cdot (-1) = \frac{3}{8} - \left(-\frac{2}{5}\right)(-1) = \frac{3}{8} - \frac{2}{5}.$$
4. With denominator $40$: $\frac{15}{40} - \frac{16}{40} = -\frac{1}{40}$.

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

"Greatest integer less than or equal to" always rounds down, toward the left on the number line. For positive numbers that looks like chopping off the decimal, but for negative numbers it moves further from zero: the result for $-0.9375$ is $-1$. As a check, this kind of remainder always has the same sign as $y$ and is smaller in size than $y$, and $-\tfrac{1}{40}$ fits both.

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