2022 AMC 10B Problem 10

Let the mean be $m$, so the median is $m + 2$ and the mode is $m + 4$. The median of five integers is the middle integer, so $m + 2$ and hence $m$ are integers.

Sort the numbers $a \le b \le c \le d \le e$, with $c = m + 2$. The mode $m + 4$ is larger than the median, so it can only occupy positions $d$ and $e$; being a mode it appears at least twice, so $d = e = m + 4$.

Since the mode is unique, no other value may appear twice. Thus $a < b < c$ (if $b = c$, the median value would also appear twice).

The sum is $5m$:
$$a + b + (m+2) + 2(m+4) = 5m \quad\Longrightarrow\quad a + b = 2m - 10.$$
With $1 \le a < b$, the smallest possible $a + b$ is $1 + 2 = 3$, so $2m - 10 \ge 3$, i.e. $m \ge 7$ (as $m$ is an integer).

For $m = 7$: $a + b = 4$, so $(a, b) = (1, 3)$ and the list is $1, 3, 9, 11, 11$. Check: mean $35/5 = 7$, median $9$, unique mode $11$. The mode is $11$.

The answer is $\boxed{\textbf{(D)}\ 11}$.

Translate each statistic into structural facts about the sorted list: the mean fixes the total, the median fixes the middle entry, and "unique mode above the median" forces the top two entries to be equal and everything else to be distinct. Minimizing then reduces to an inequality on the two smallest entries. Always check the integrality that the median of an odd-sized integer list provides.

Test the choices from the bottom. Mode $5$ means mean $1$, impossible with five positive integers summing to $5$ unless all are $1$. Mode $7$ or $9$ means mean $3$ or $5$ with the top two entries $7, 7$ or $9, 9$, and the remaining sum for $a + b$ comes out negative or below $3$. Mode $11$ works with $1, 3, 9, 11, 11$.

• Letting the two smallest numbers be equal (which creates a second mode) or forgetting that the mean must be an integer here.
• Placing the mode at the median position or below it, even though the mode is stated to be larger than the median.