2021 AMC 10A Problem 16

1. Count the entries: $1 + 2 + \cdots + 200 = \frac{200 \times 201}{2} = 20100$. That is even, so the median is the average of the $10050$th and $10051$st entries. The list is already in order.
2. Look at how the first few numbers fill positions:

The last position filled by $n$ is $1 + 2 + \cdots + n = \frac{n(n+1)}{2}$.
3. We want the first $n$ whose last position reaches $10050$, so $n(n+1)$ should be about $20100$. Since $n \times n$ is close to $n(n+1)$, try $n$ near the square root of $20100$. We have $141^2 = 19881$ and $142^2 = 20164$, so check $141$ and $142$.
4. $\frac{141 \times 142}{2} = 10011$, so the numbers $1$ to $141$ end at position $10011$.
5. $\frac{142 \times 143}{2} = 10153$, so the number $142$ fills positions $10012$ through $10153$.
6. Both $10050$ and $10051$ are in that stretch, so both middle entries are $142$. The median is $142$.

The answer is $\boxed{\textbf{(C)}\ 142}$.

For a list made of blocks of repeated numbers, first find the middle position, then find which block contains it by counting how many positions the earlier blocks use up. Here the running count is a triangular number, $\frac{n(n+1)}{2}$, and a square root tells you where to look. Always check that both middle positions fall in the same block.

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 $\boxed{\textbf{(C)}\ 142}$.