2024 AMC 10B Problem 15

Turn each fact into a rule, then search case by case.

Rule from the mean.

1. The six known numbers add to $1 + 2.2 + 3.2 + 5.2 + 6.2 + 7 = 24.8$.
2. The mean is a whole number, so the total of all nine is a multiple of $9$: $27, 36, 45, \ldots$ So $x + y + z$ is $2.2$, $11.2$, $20.2$, and so on.
3. The range is $7$, and the list already contains $1$ and $7$, so every number is between $0$ and $8$. Then $x + y + z \le 24$, which leaves $2.2$, $11.2$ or $20.2$.

Rule from the median.

1. The median is the $5$th of the $9$ numbers in order, and it must be a whole number. The number $1$ is too small to be $5$th and $7$ too big, and $2.2, 3.2, 5.2, 6.2$ are not whole numbers. So the median is one of $x, y, z$.

Rule from the range. The smallest number is $x$ or $1$, and the largest is $z$ or $7$. That gives four cases.

Case 1: smallest $x$, largest $7$. Then $7 - x = 7$, so $x = 0$, and $z \le 7$.

5. Now $y + z$ is $2.2$, $11.2$ or $20.2$. It cannot be $20.2$, because $y$ and $z$ are each at most $7$. It cannot be $2.2$: then $y$ and $z$ are both at most $2.2$, the five smallest numbers are $0, 1, y, z, 2.2$, and the $5$th is $2.2$, not a whole number. So $y + z = 11.2$, and since $z \le 7$, $y \ge 4.2$.
The list in order starts $0, 1, 2.2, 3.2$. If $y$ is more than $5.2$, the $5$th number is $5.2$, which is not whole. So $y$ is at most $5.2$ and it is the $5$th number.
6. $y$ must be a whole number, so $y = 5$ and $z = 6.2$. Triple: $(0, 5, 6.2)$.

Case 2: smallest $x$, largest $z$. Then $z = x + 7$ and $x \le 1$.

7. The list starts $x, 1, 2.2, 3.2$, so the $5$th number is $y$ if $y$ is between $3.2$ and $5.2$. So $y = 4$ or $y = 5$.
8. Total: $x + y + (x + 7) = 2x + y + 7$ must be $11.2$ or $20.2$.
- $y = 4$: $2x = 0.2$ gives $x = 0.1$; $2x = 9.2$ gives $x = 4.6$, too big. Triple: $(0.1, 4, 7.1)$.
- $y = 5$: $2x = -0.8$ or $2x = 8.2$; neither gives $x$ between $0$ and $1$.

Case 3: smallest $1$, largest $z$. Then $z = 8$ and $x \ge 1$.

9. $x + y$ is $3.2$ or $12.2$.
- $x + y = 3.2$: both are at most $2.2$, so the list in order is $1, x, y, 2.2, 3.2, \ldots$ and the $5$th number is $3.2$. Not a whole number.
- $x + y = 12.2$: since $y \le 8$, $x \ge 4.2$. If $x < 5.2$, the $5$th number is $5.2$. If $x > 6.2$, then $y < 6$, which is smaller than $x$, impossible. So $x$ is between $5.2$ and $6.2$, and $x$ is the $5$th number, so $x = 6$ and $y = 6.2$. Triple: $(6, 6.2, 8)$.

Case 4: smallest $1$, largest $7$. The range is $6$, not $7$. No triples.

In total there are three triples: $(0, 5, 6.2)$, $(0.1, 4, 7.1)$, $(6, 6.2, 8)$.

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

Each piece of data becomes a concrete rule. The mean fixes what $x + y + z$ can be. The median says one of $x, y, z$ sits in the middle as a whole number. The range says which numbers are at the two ends. Splitting into cases by the two ends keeps the search short, and in each case the other rules decide the rest.