2012 AMC 10A Problem 13

Suppose the numbers are entered in the order $p, q, r, s, t$. Track the coefficient of each one:
$$\frac{p + q}{2} \;\to\; \frac{p + q}{4} + \frac{r}{2} \;\to\; \frac{p + q}{8} + \frac{r}{4} + \frac{s}{2} \;\to\; \frac{p + q}{16} + \frac{r}{8} + \frac{s}{4} + \frac{t}{2}.$$
Each averaging step halves every earlier weight, so later entries count more. The result is a weighted average with weights $\frac1{16}, \frac1{16}, \frac18, \frac14, \frac12$.

To maximize, give the biggest weights to the biggest numbers: order $1, 2, 3, 4, 5$ gives
$$\frac{1 + 2}{16} + \frac{3}{8} + \frac{4}{4} + \frac{5}{2} = \frac{3 + 6 + 16 + 40}{16} = \frac{65}{16}.$$
To minimize, reverse: order $5, 4, 3, 2, 1$ gives
$$\frac{5 + 4}{16} + \frac{3}{8} + \frac{2}{4} + \frac{1}{2} = \frac{9 + 6 + 8 + 8}{16} = \frac{31}{16}.$$

Difference: $\frac{65 - 31}{16} = \frac{34}{16} = \frac{17}{8}$.

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

Repeated averaging is linear, so the output is a fixed weighted average of the inputs; only the weights depend on position. Once the weights are known, extremes come from the rearrangement principle: pair large weights with large values for the maximum and with small values for the minimum. Whenever a process "averages with the next number," expect the weights $\frac12, \frac14, \frac18, \dots$

• Assuming the answer is just 5 - 1 = 4 or computing the max and min with the wrong orders (e.g. 1,2,3,4,5 ascending gives the max, not descending).
• Giving the first two entries weights $\frac12$ and $\frac14$ instead of realizing they share the smallest weight $\frac1{16}$.