Question

A list of $9$ real numbers consists of $1,$ $2.2,$ $3.2,$ $5.2,$ $6.2,$ $7,$ as well as $x, y, z$ with $x \le y \le z.$ The range of the list is $7,$ and the mean and median are both positive integers. How many ordered triples $(x, y, z)$ are possible?

$1$

$2$

$3$

$4$

infinitely many

Solution:

The six fixed numbers total $24.8,$ so an integer mean needs $x + y + z = 9k - 24.8$ for some positive integer $k.$ A range of $7$ pins down the overall min and max (the fixed values already stretch from $1$ to $7$ ), and the median is the $5$ th smallest of the nine numbers, which has to be a positive integer. Grind through where $x, y, z$ can sit and exactly three triples survive: $(0, 5, 6.2),$ $(0.1, 4, 7.1),$ and $(6, 6.2, 8).$ So there are $3.$ Thus, C is the correct answer.

2024 AMC 10B Problem 15

Solution:

Problem 15 in Other Years