2024 AMC 10B Problem 15

The six given numbers sum to $24.8$, so an integer mean needs $x + y + z \in \{2.2, 11.2, 20.2, \dots\}$. Range $7$ puts every value in $[0, 8]$, so only $2.2$, $11.2$, $20.2$ are possible.

Median: the fifth smallest of nine values must be an integer. Neither $1$ nor $7$ can be fifth, and the other given numbers are not integers, so the median is one of $x, y, z$.

Split by which values are extreme.

Min $x$, max $7$. Then $x = 0$, $z \le 7$. The four smallest are $0, 1, 2.2, 3.2$, so $y$ is the median with $y \in \{4, 5\}$, and $y + z = 11.2$. $y = 4$ gives $z = 7.2 > 7$; $y = 5$ gives $z = 6.2$. Triple $(0, 5, 6.2)$.

Min $x$, max $z$. Then $z = x + 7$, $x \le 1$, and again $y \in \{4, 5\}$. From $2x + y + 7 \in \{11.2, 20.2\}$: $y = 4$ gives $x = 0.1$ (or $4.6$, too big); $y = 5$ gives nothing valid. Triple $(0.1, 4, 7.1)$.

Min $1$, max $z$. Then $z = 8$, $x \ge 1$, and $x + y \in \{3.2, 12.2\}$. Sum $3.2$ leaves $3.2$ as median, rejected. For $12.2$, $x \ge 4.2$; the fifth value is $5.2$ if $x < 5.2$ and $6.2$ if $x > 6.2$, so $x$ is the median with $x = 6$, $y = 6.2$. Triple $(6, 6.2, 8)$.

Min $1$, max $7$ has range $6$, impossible.

Three triples. The answer is $\boxed{\textbf{(C)}\ 3}$.

Each statistic translates into a concrete constraint: the mean into a divisibility condition on the sum, the range into "who is the extreme value," and the median into "which unknown is fifth and is an integer." Organizing by the range (four cases on the min and max) keeps the search finite and forces one unknown per case, after which the sum condition determines the rest.

• Treating the median as 5.2 or one of the given numbers; with nine values the fifth must be an integer, so one of x, y, z has to sit in the middle.
• Accepting $(0, 4, 7.2)$ without checking the range; $7.2 - 0 > 7$ so it fails.