2022 AMC 10B Problem 10

Give the mean a name and write everything else in terms of it. Say the mean is $m$. Then the median is $m + 2$ and the mode is $m + 4$.

1. The mean is a whole number. With five numbers, the median is the middle one, so it is a whole number. The mean is $2$ less, so it is a whole number too.
2. Where the mode sits. List the numbers from smallest to largest: $a, b, c, d, e$. The median is $c = m + 2$. The mode is bigger than $c$, so it can only be $d$ or $e$. A mode appears at least twice, so $d = e = m + 4$.
3. No other repeats. The mode is the only one, so no other value may appear twice. So $a$, $b$ and $c$ are all different, which means $a < b < c$.
4. Use the total. Five numbers with mean $m$ add up to $5m$:
$$a + b + (m + 2) + (m + 4) + (m + 4) = 5m,$$
which simplifies to $a + b = 2m - 10$.
5. Smallest $m$ (cannot go lower). The smallest two different positive whole numbers are $1$ and $2$, so $a + b \ge 3$. Then $2m - 10 \ge 3$, so $m \ge 6.5$. Since $m$ is a whole number, $m \ge 7$.
6. $m = 7$ works. Then $a + b = 4$ with $a < b$, so $a = 1$, $b = 3$. The list is $1, 3, 9, 11, 11$. Check: the sum is $35$ and the mean is $7$; the median is $9$; the only repeated value is $11$. Everything fits.
7. The mode is $m + 4 = 11$.

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

Turn each word into a fact about the sorted list: the mean fixes the total, the median is the middle entry, and a single mode above the median has to take the top two spots. After that, making the answer small is just a matter of making the two smallest entries as small as the rules allow. Do not forget that the median of an odd number of whole numbers is itself a whole number.

Test the answer choices from the smallest. Mode $5$ means mean $1$, but five positive whole numbers with mean $1$ are all $1$, so the mode would be $1$. Mode $7$ means mean $3$ and total $15$; the numbers $5, 7, 7$ already add to $19$, too much. Mode $9$ means mean $5$ and total $25$; the numbers $7, 9, 9$ add to $25$, leaving $0$ for $a + b$, impossible. Mode $11$ works with $1, 3, 9, 11, 11$.

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 $\boxed{\textbf{(D)}\ 11}$.