2003 AMC 8 Problems/Problem 23

Revision as of 19:47, 20 November 2020 by Srijankundu (talk | contribs) (Problem)

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Solution

Break this problem into two parts: where the cat will be after the $247^{th}$ move, and where the mouse will be.

The cat has four possible configurations which are repeated every four moves. $247$ has a remainder of $3$ when divided by $4$. This corresponds to the position the cat has after the 3rd move, which is the bottom right corner.

Similarly, the mouse has eight possible configurations that repeat every eight moves. $247$ has a remainder of $7$ when divided by $8$. This corresponds to the position the rat has after the 7th move, which can easily be found by writing two more steps to be the bottom edge on the left side of the grid.

The only configuration with the mouse in that position and the cat in the bottom right square is $\boxed{\textbf{(A)}}$.

Video Solution

https://youtu.be/RCUzhVOi7XI ~DSA_Catachu

See Also

2003 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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