2019 AMC 10A Problem 10

Think about what happens as the bug walks: it stays in one tile until it crosses a grid line, and then it is in a new tile. So

$$\text{tiles visited} = 1 \text{ (the starting tile)} + \text{number of grid lines crossed},$$

as long as the bug never crosses two lines at the same moment (by walking exactly through a corner where four tiles meet).

1. Lines across the length. The floor is $17$ feet long, so going along the length there are $17 - 1 = 16$ grid lines between the end walls. The bug crosses each one once.
2. Lines across the width. The floor is $10$ feet wide, so there are $10 - 1 = 9$ grid lines between the side walls. The bug crosses each one once.
3. Does it ever hit a corner? Put the start at $(0,0)$ and the end at $(17,10)$. A point on the path is $17$ parts across for every $10$ parts up. To hit a tile corner, both coordinates must be whole numbers at once, and since $17$ and $10$ have no common factor other than $1$, that only happens at the start and the end. So all crossings happen at different moments.
4. Count. $16 + 9 = 25$ crossings, so the bug visits $1 + 25 = 26$ tiles.

The answer is $\boxed{\textbf{(C)}\ 26}$.

"New tile each time you cross a line" turns a picture into a count. The one thing that could spoil it is walking exactly through a corner, where two crossings happen at once and only one new tile is entered. For a floor $m$ by $n$ whose sides share no common factor, that never happens and the answer is $m + n - 1$. If they do share a factor, subtract one tile for each corner the path passes through.

Check the rule on a tiny floor you can draw: a $2 \times 3$ floor. The diagonal crosses $2$ lines one way and $1$ the other, and visits $1 + 3 = 4$ tiles, which you can confirm by sketching. Using the same rule on $10 \times 17$ gives $10 + 17 - 1 = 26$.