2024 AMC 10B Problem 15

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

Answer: C

Concepts:mean median (data)range casework

Difficulty rating: 1730

Small hint:

The six given numbers sum to

24.8,

so

x + y + z

must make the total a multiple of

9

Big hint:

Range

7

fixes the smallest and largest values; then require the

5

th smallest number to be a positive integer

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

to

7,

the range condition gives three cases.

If

z\le7,

then

x=0.

The bounds on

y+z

force

k=3

or

4.

For

k=3,

the median is

2.2;

for

k=4,

we have

y+z=11.2,

so

y\ge4.2

and the median is integral only when

y=5,

giving

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

If

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,

so

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.

2024 AMC 10B Problem 15

Answer: C

Difficulty rating: 1730

Small hint:

Big hint:

Solution:

Problem 15 in other years