2002 AMC 10A Problem 22

Only the number of tiles matters, since renumbering restores the set $\{1, \dots, N\}$. With $N$ tiles, the operation deletes the $\lfloor \sqrt{N} \rfloor$ perfect squares up to $N$.

Follow what happens starting from a perfect square, $N = n^2$:

• First operation: $n$ squares are removed, leaving $n^2 - n$ tiles.
• Since $(n-1)^2 = n^2 - 2n + 1 < n^2 - n < n^2$, the largest square not exceeding $n^2 - n$ is $(n-1)^2$, so the second operation removes $n - 1$ tiles, leaving $n^2 - n - (n - 1) = n^2 - 2n + 1 = (n-1)^2$.

So every two operations move from $n^2$ tiles to $(n-1)^2$ tiles. Starting at $100 = 10^2$, reaching $1 = 1^2$ requires the nine transitions $10 \to 9 \to \dots \to 1$, i.e. $2 \cdot 9 = 18$ operations.

Concretely: $100 \to 90 \to 81 \to 72 \to 64 \to 56 \to 49 \to \dots \to 4 \to 2 \to 1$.

The answer is $\boxed{\textbf{(C)}\ 18}$.

Iterated processes often have a hidden invariant: here, after every second step the tile count is again a perfect square, and the square root drops by exactly one. Discovering it takes only two or three hand computations ($100, 90, 81, 72, 64$), after which the structure is clear and the count $2(10 - 1)$ follows. When the answer choices spread out ($10$, $11$ versus $18$, $19$, $20$), the pattern must be nailed down precisely, including whether the final step is counted.

Just tabulate: the removed counts are $10, 9, 9, 8, 8, 7, 7, 6, 6, 5, 5, 4, 4, 3, 3, 2, 2, 1$, whose running totals bring $100$ down to $1$. The value $10$ appears once, each of $9$ down to $2$ twice, and $1$ once (the final step from $2$ tiles to $1$): $1 + 16 + 1 = 18$ entries.

• Assuming each operation removes 10 tiles (answer 10) or that the count drops one square number per step (answer 9 or 10).
• Off-by-one at the end: stopping when the count reaches a single square ($4$ tiles after $16$ steps) or counting an extra step after already reaching $1$ tile, giving $19$.