2019 AMC 10B Problem 13

The mean is $\frac{4 + 6 + 8 + 17 + x}{5} = \frac{35 + x}{5}$. The median depends on where $x$ sits relative to $6$ and $8$.

Case $x \le 6$: the ordered list is $x, 4, 6, 8, 17$ or $4, x, 6, 8, 17$, so the median is $6$. Then $\frac{35 + x}{5} = 6$ gives $x = -5$, which is indeed $\le 6$. Valid.

Case $6 < x < 8$: the median is $x$ itself. Then $\frac{35 + x}{5} = x$ gives $35 = 4x$, $x = \frac{35}{4} = 8.75$, which is not between $6$ and $8$. Rejected.

Case $x \ge 8$: the median is $8$. Then $\frac{35 + x}{5} = 8$ gives $x = 5$, which is not $\ge 8$. Rejected.

The only solution is $x = -5$, so the sum of all solutions is $-5$.

The answer is $\boxed{\textbf{(A)}\ -5}$.

The median is a piecewise function of $x$, so the equation "median equals mean" is really three linear equations, each valid only on its own interval. Solving each and then checking membership in the interval is the whole method; two of the three candidates fail that check. Always verify a casework solution against the assumption that generated it.

• Adding all three candidate values (-5, 5 and 35/4) without checking that 5 and 35/4 violate the case that produced them.
• Taking the median of the four given numbers ($7$) and ignoring that $x$ changes the ordering.