2012 AMC 12A Problem 8

Question

An iterative average of the numbers

1,

2,

3,

4,

and

5

is computed in the following way. Arrange the five numbers in some order. Find the mean of the first two numbers, then find the mean of that with the third number, then the mean of that with the fourth number, and finally the mean of that with the fifth number. What is the difference between the largest and smallest possible values that can be obtained using this procedure?

\dfrac{31}{16}

2

\dfrac{17}{8}

3

\dfrac{65}{16}

Answer: C

Concepts:mean optimization

Difficulty rating: 1480

Small hint:

Work out the weight each position contributes to the final average

Big hint:

For the order

a,b,c,d,e

the result is

\dfrac{a+b+2c+4d+8e}{16}

; the last entries matter most

Solution:

For the order

a, b, c, d, e,

the iterative average is

\frac{a + b + 2c + 4d + 8e}{16}.

The later positions carry the most weight.

The largest value uses

(a,b,c,d,e) = (1,2,3,4,5),

giving

\dfrac{65}{16},

and the smallest uses

(5,4,3,2,1),