2018 AMC 10B Problem 14

We want as few different values as possible, so each value should be used as many times as the rules allow.

1. The mode's spots. The mode appears exactly $10$ times. That leaves $2018 - 10 = 2008$ other spots.
2. The limit on other values. The mode is the only most common value, so every other value appears fewer than $10$ times, that is, at most $9$ times.
3. Fewest other values. Each other value fills at most $9$ spots. Divide:
$$2008 \div 9 = 223 \text{ remainder } 1.$$
So $223$ values fill $223 \times 9 = 2007$ spots, and $1$ spot is still empty. That needs one more value. So at least $224$ other values are needed.
4. It can be done. Use $223$ values that appear $9$ times each, plus one more value that appears once. Together with the mode, that fills all $2018$ spots, and the mode is still the only value that appears $10$ times.
5. Count everything. $224$ other values plus the mode itself gives $224 + 1 = 225$.

The answer is $\boxed{\textbf{(D)}\ 225}$.

To use the fewest values, pack each value as full as the rules allow. The rule "unique mode" caps every other value at $9$ uses. When the division leaves a remainder, you need one more value to cover it. And the question asks for all the values, so do not forget the mode.