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Difference between revisions of "2003 AMC 8 Problems/Problem 23"

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==Problem==
 
==Problem==
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In the pattern below, the cat (denoted as a large circle in the figures below) moves clockwise through the four squares  and the mouse (denoted as a dot in the figures below) moves counterclockwise through the eight exterior segments of the four squares.
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In the pattern below, the cat moves clockwise through the four squares, and the mouse moves counterclockwise through the eight exterior segments of the four squares.
  
 
<center>
 
<center>
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==Solution==
 
==Solution==
  
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We'll break down this problem into two parts: <math>1)</math> where the cat will be after the 247th move and <math>2)</math> where the mouse will be after the 247th move.  
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Break this problem into two parts: where the cat will be after the <math>247^{th}</math> move, and where the mouse will be.
  
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Instead of using formulas two solve this problem we'll just look for patterns. Let's start off with the cat. After 4 rotations clockwise the cat will be back in its starting position. So the cat will be back in its starting position in the 5th move. Similarly, we can find that the cat will be in its starting position if the number of moves <math>n</math> is divisible by 5. By dividing 247 by 5 we get a remainder of 2. So after two more moves the cat will be in the bottom right square.
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The cat has four possible positions in 1 cycle which are repeated every four moves. <math>247</math> has a remainder of <math>3</math> when divided by <math>4</math>. This corresponds to the position the cat has after the 3rd move, which is the bottom right corner.
  
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Similarly, we can find that the mouse will be back on its starting position for every move <math>n</math> which is divisible by 9. Dividing 247 by 9 we get a remainder of 5. After 5 moves the mouse will be on the top segment of the left hand side of the big square.  
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Similarly, the mouse has eight possible positions in 1 cycle that repeat every eight moves. <math>247</math> has a remainder of <math>7</math> when divided by <math>8</math>. This corresponds to the position the rat has after the 7th move which is bottom left corner.  
  
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So the answer is <math>\boxed{\bolded{\text{C}}}</math>.
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The only arrangement with the mouse in that position and the cat in the bottom right square is <math>\boxed{\textbf{(A)}}</math>.
  
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==Video Solution==
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https://youtu.be/RCUzhVOi7XI
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~DSA_Catachu
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https://www.youtube.com/watch?v=OxtaQkcJDfU
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==Video Solution #2==
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https://youtu.be/xMy3wMFF3KQ Soo, DRMS, NM
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==See Also==
 
{{AMC8 box|year=2003|num-b=22|num-a=24}}
 
{{AMC8 box|year=2003|num-b=22|num-a=24}}
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{{MAA Notice}}

Revision as of 22:24, 4 November 2022

Problem

In the pattern below, the cat moves clockwise through the four squares, and the mouse moves counterclockwise through the eight exterior segments of the four squares.

2003amc8prob23a.png

If the pattern is continued, where would the cat and mouse be after the 247th move?

2003amc8prob23b.png

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 positions in 1 cycle 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 positions in 1 cycle 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 is bottom left corner.

The only arrangement 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

https://www.youtube.com/watch?v=OxtaQkcJDfU

Video Solution #2

https://youtu.be/xMy3wMFF3KQ Soo, DRMS, NM


See Also

2003 AMC 8 (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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