2000 AMC 10 Problem 23

1. The mode. The number $2$ appears three times, and adding one $x$ cannot give any other number three appearances. So the mode is always $2$.
2. The mean. The numbers other than $x$ add to $10 + 2 + 5 + 2 + 4 + 2 = 25$, so the mean is $\frac{25 + x}{7}$.
3. The median. Without $x$, the sorted list is $2, 2, 2, 4, 5, 10$. With $x$ there are $7$ numbers, and the median is the $4$th one. So:
- if $x \le 2$, the median is $2$;
- if $x$ is between $2$ and $4$, the median is $x$;
- if $x \ge 4$, the median is $4$.
4. "Arithmetic progression" means the three values, in order, are evenly spaced, like $2, 3, 4$. "Non-constant" means they are not all equal. Now check each case for the median.

Case 1: median $2$. Then the mode and the median are both $2$. Evenly spaced numbers with two equal would all be equal, which is not allowed. No solutions.

Case 2: median $4$ ($x \ge 4$). We have $2$, $4$ and the mean. The mean can go in three places:
- after them: $2, 4, 6$, so the mean is $6$, giving $25 + x = 42$ and $x = 17$;
- between them: $2, 3, 4$, so the mean is $3$, giving $x = -4$;
- before them: $0, 2, 4$, so the mean is $0$, giving $x = -25$.

Only $x = 17$ fits $x \ge 4$. Check: the mean is $\frac{42}{7} = 6$, the median is $4$, the mode is $2$. That is $2, 4, 6$. It works.

Case 3: median $x$ (between $2$ and $4$). We have $2$, $x$ and the mean, with $2$ smaller than $x$.
- Order $2, x, \text{mean}$: the middle one is the average of the other two, so $2x = 2 + \frac{25 + x}{7}$. Multiply by $7$: $14x = 14 + 25 + x$, so $13x = 39$ and $x = 3$. The mean is $\frac{28}{7} = 4$, giving $2, 3, 4$. It works.
- Order $2, \text{mean}, x$: $2 \cdot \frac{25 + x}{7} = 2 + x$ gives $50 + 2x = 14 + 7x$, so $x = \frac{36}{5} = 7.2$. That is not between $2$ and $4$.
- Order $\text{mean}, 2, x$: $4 = \frac{25 + x}{7} + x$ gives $28 = 25 + 8x$, so $x = \frac38$. Not between $2$ and $4$.

1. The possible values are $x = 17$ and $x = 3$. Their sum is $20$.

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

Two of the three statistics are easy: the mode never changes and the mean is a simple formula. Only the median depends on where $x$ lands, so split into cases by the median. Each case gives a simple equation, and every answer must be checked against the range of $x$ that defined its case. The word "non-constant" rules out the whole first case.

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 $\boxed{\textbf{(E)}\ 20}$.