2021 AMC 10A Problem 16

The list has $1 + 2 + \dots + 200 = \frac{200 \cdot 201}{2} = 20100$ entries, an even number, so the median is the average of the $10050$th and $10051$st entries (the list is already sorted).

The values $1, 2, \dots, n$ together fill the first $\frac{n(n+1)}{2}$ positions. We want the smallest $n$ with $\frac{n(n+1)}{2} \ge 10050$, i.e. $n(n+1) \ge 20100$. Since $n^2 \approx 20100$ suggests $n \approx 142$:
$$\frac{141 \cdot 142}{2} = 10011 < 10050,\qquad \frac{142 \cdot 143}{2} = 10153 \ge 10051.$$
Positions $10012$ through $10153$ all hold the value $142$, which includes both middle positions.

The median is $\frac{142 + 142}{2} = 142$, so the answer is $\boxed{\textbf{(C)}\ 142}$.

For a list built from blocks, locate the middle position, then find which block contains it using cumulative counts. Triangular numbers give the cumulative count in closed form, and $\sqrt{2 \cdot 10050} \approx 141.8$ tells you exactly where to check. Verify that both middle positions land in the same block before declaring the median an integer.

• Taking the median of the distinct values 1 to 200 (100.5), or averaging the values 141 and 142 after misplacing the middle positions.
• Solving $\frac{n(n+1)}{2} = 10050$ and rounding down to $141$ without checking that $10011 < 10050$.