2000 AMC 10 Problem 23

The value $2$ occurs three times and nothing else can catch up, so the mode is $2$ regardless of $x$. The mean is $\frac{25 + x}{7}$. Sorted without $x$ the list is $2, 2, 2, 4, 5, 10$; the median is the fourth of seven values, so it is $2$ if $x \le 2$, it is $x$ if $2 \le x \le 4$, and it is $4$ if $x \ge 4$.

Median $2$ ($x \le 2$). Mode and median both equal $2$; a progression with a repeated term is constant. Rejected.

Median $4$ ($x \ge 4$). The three numbers are $2$, $4$ and the mean. For them to form a progression, the mean must be $6$ (giving $2, 4, 6$), $3$ (giving $2, 3, 4$) or $0$ (giving $0, 2, 4$). Solving $\frac{25 + x}{7} = 6, 3, 0$ gives $x = 17, -4, -25$. Only $17$ satisfies $x \ge 4$.

Median $x$ ($2 < x < 4$). The numbers are $2$, $x$ and the mean, with $2 < x$.
- Order $2, x, \text{mean}$: $2x = 2 + \frac{25 + x}{7}$, so $14x = 39 + x$ and $x = 3$. Then the mean is $4$ and $2, 3, 4$ works.
- Order $2, \text{mean}, x$: $\frac{2(25 + x)}{7} = 2 + x$ gives $x = \frac{36}{5}$, outside the range.
- Order $\text{mean}, 2, x$: $\frac{25 + x}{7} = 4 - x$ gives $x = \frac38$, outside the range.

The possible values are $17$ and $3$, with sum $20$.

The answer is $\boxed{\textbf{(E)}\ 20}$.

Two of the three statistics are pinned down (the mode outright, the mean as a formula), and the third, the median, is piecewise in $x$. Casework on the median's piece converts the problem into a handful of linear equations, each of which must be checked against the range that defined its case. "Non-constant" is not decoration; it kills the entire first case.

• Assuming the mean is always the largest of the three statistics; when the median is x you must test all three orderings, and when the median is 4 you must discard solutions that leave the case's range.
• Keeping $x = -4$ or $x = -25$ from the median-$4$ case; for those values the median would actually be $2$.
• Overlooking the word "non-constant" and admitting $x = -11$ (which makes mean, median and mode all $2$).