2024 AMC 12B Problem 10

Problem 10 of 25EasierAlgebra

A list of

9

real numbers consists of

1,

2.2,

3.2,

5.2,

6.2,

and

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 sum to

24.8

and span

[1,7].

There are three possible arrangements of the overall extremes. If the extremes are

0

and

7,

then

x=0

and

z\le7.

The only possible integer means are

3

and

4,

requiring

y+z=2.2

11.2.

The first makes the median

2.2.

In the second, an integer median forces

y=5,

hence

z=6.2.

This gives

(0,5,6.2).

If the extremes are

1

and

8,

then

z=8

and the integer mean forces

x+y=3.2

12.2.

The first makes the median

3.2;

the second has an integer median only for

x=6,\ y=6.2.

This gives

(6,6.2,8).

Finally, if both extremes are new, write

x=t,\ z=t+7

with

0\lt t\lt1.

The mean must be

4,

y=4.2-2t.

The median is the fourth number among

1,2.2,3.2,5.2,6.2,7,y;

it is an integer only when

y=4,

giving

t=0.1.

Thus the third triple is

(0.1,4,7.1),

and there are exactly

3

in all. Thus, the correct answer is C.