2020 AMC 10B Problem 15

Nobody can erase $10{,}000$ digits by hand, so look for a pattern that repeats.

1. Pick a good block size. The original list repeats every $5$ digits. The first erasure removes every $3$rd digit, so take a block whose length is a multiple of both $5$ and $3$. We use $30$, because it makes the later steps line up too:
- $30$ digits, erase every $3$rd, leaves $20$ digits;
- $20$ is a multiple of $4$: erase every $4$th, leaves $15$;
- $15$ is a multiple of $5$: erase every $5$th, leaves $12$.

Each block length is a multiple of the next step size, so every block is treated in exactly the same way. The final list therefore repeats every $12$ digits.
2. Do one block by hand. The first $30$ digits are $12345$ written six times.
- Erase every $3$rd digit: $1\,2\,4\,5\,2\,3\,5\,1\,3\,4\,1\,2\,4\,5\,2\,3\,5\,1\,3\,4$ ($20$ digits).
- Erase every $4$th of those: $1\,2\,4\,2\,3\,5\,3\,4\,1\,4\,5\,2\,5\,1\,3$ ($15$ digits).
- Erase every $5$th of those: $1\,2\,4\,2\,5\,3\,4\,1\,5\,2\,5\,1$ ($12$ digits).
3. Find positions 2019 to 2021. Divide by $12$: $2019 = 12 \times 168 + 3$. So position $2019$ is position $3$ of a block, and $2020$, $2021$ are positions $4$ and $5$.
4. Positions $3, 4, 5$ of the block are $4, 2, 5$. Their sum is $4 + 2 + 5 = 11$.
5. Is the list long enough? $10{,}000$ digits become about $6667$, then $5001$, then $4001$. That is more than $2021$, so those positions exist.

The answer is $\boxed{\textbf{(D)}\ 11}$.

Erasing every $k$th item from a repeating list gives another repeating list. Choose a block length that fits the old pattern and the erasing step, and every block is treated the same way. Working one block by hand and then dividing by the block length replaces thousands of steps with one small calculation.