2012 AMC 10A Problem 13

First see how much each position "counts", then choose the best and worst orders.

1. Try the order $1, 2, 3, 4, 5$ to see the pattern:
- mean of $1$ and $2$: $1.5$,
- mean of $1.5$ and $3$: $2.25$,
- mean of $2.25$ and $4$: $3.125$,
- mean of $3.125$ and $5$: $4.0625 = \frac{65}{16}$.
2. Now the general pattern. Each time we take a mean, the new number gets half the weight, and everything before it is cut in half. Following the five positions through:
- after the last step, the fifth number counts for $\frac12$,
- the fourth for $\frac14$,
- the third for $\frac18$,
- the first and second for $\frac1{16}$ each.
These add to $1$, so the result is a weighted average: later numbers count more.
3. Largest result. Put the biggest numbers where they count most, so the order is $1, 2, 3, 4, 5$. As in step 1:
$$\frac{1 + 2}{16} + \frac{3}{8} + \frac{4}{4} + \frac{5}{2} = \frac{3 + 6 + 16 + 40}{16} = \frac{65}{16}.$$
4. Smallest result. Put the smallest numbers where they count most, so the order is $5, 4, 3, 2, 1$:
$$\frac{5 + 4}{16} + \frac{3}{8} + \frac{2}{4} + \frac{1}{2} = \frac{9 + 6 + 8 + 8}{16} = \frac{31}{16}.$$
5. Difference: $\frac{65}{16} - \frac{31}{16} = \frac{34}{16} = \frac{17}{8}$.

The answer is $\boxed{\textbf{(C)}\ \frac{17}{8}}$.

Repeated averaging always gives each position a fixed share of the result: $\frac12, \frac14, \frac18, \dots$ counting back from the end. Once you know the shares, the largest result comes from giving the biggest share to the biggest number, and the smallest result from giving it to the smallest number.