2024 AMC 10B Problem 15

Problem 15 of 25IntermediateAlgebra

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?

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Solution

The six fixed numbers total

24.8.

If the mean is the integer

k,

then

x+y+z=9k-24.8.

Because the fixed entries already run from

1

7,

the range condition gives three cases. If

z\le7,

then

x=0.

The bounds on

y+z

force

k=3

4.

For

k=3,

the median is

2.2;

for

k=4,

we have

y+z=11.2,

y\ge4.2

and the median is integral only when

y=5,

giving

(x,y,z)=(0,5,6.2).

x\ge1,

then

z=8.

Here

k=4

gives median

3.2.

For

k=5,

we have

x+y=12.2;

the median is integral only when

x=6,

giving

(6,6.2,8).

The remaining case has

0<x<1

and

z=x+7.

The total lies between

31.8

and

41.8,

k=4

and

y=4.2-2x.

The median can be an integer only when

y=4,

giving

(0.1,4,7.1).

Hence exactly

3

ordered triples work. Thus, C is the correct answer.