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Difference between revisions of "2024 AMC 8 Problems/Problem 17"

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==Solution 1==
 
==Solution 1==
  
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Corners have <math>5</math> spots to go and <math>4</math> corners so <math>5 \times 4=20</math>.
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Corners have <math>5</math> spots to go and there are <math>4</math> corners, so <math>5 \times 4=20</math>.
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Edges have <math>3</math> spots to go and <math>4</math> sides so <math>3 \times 4=12</math>
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Edges have <math>3</math> spots to go and there are <math>4</math> sides so, <math>3 \times 4=12</math>.
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<math>20+12=32</math> in total.
+
That gives us <math>20+12=32</math> spots to go into totally.
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<math>\boxed{\textbf{(E)} 32}</math> is the answer.
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So <math>\boxed{\textbf{(E)} 32}</math> is the answer.
  
 
~Evergreen Middle School
 
~Evergreen Middle School

Revision as of 18:29, 30 January 2024

Problem

A chess king is said to attack all the squares one step away from it, horizontally, vertically, or diagonally. For instance, a king on the center square of a $3$ x $3$ grid attacks all $8$ other squares, as shown below. Suppose a white king and a black king are placed on different squares of a $3$ x $3$ grid so that they do not attack each other. In how many ways can this be done?

$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 27 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 32$

Solution 1

Corners have $5$ spots to go and there are $4$ corners, so $5 \times 4=20$. Edges have $3$ spots to go and there are $4$ sides so, $3 \times 4=12$. That gives us $20+12=32$ spots to go into totally. So $\boxed{\textbf{(E)} 32}$ is the answer.

~Evergreen Middle School

Solution 2

We see that the center is not a viable spot for either of the kings to be in, as it would attack all nearby squares.

This gives three combinations:

Corner-corner: There are 4 corners, and none of them are touching orthogonally or diagonally, so it's $\binom{4}{2}=6$

Corner-edge: For each corner, there are two edges that don't border it, $4\cdot2=8$

Edge-edge: The only possible combinations of this that work are top-bottom and left-right edges, so $2$ for this type


$6+8+2=16$

Multiply by two to account for arrangements of colors to get $\fbox{E) 32}$ ~ c_double_sharp

Video Solution 1 (super clear!) by Power Solve

https://youtu.be/SG4PRARL0TY

Video Solution 2 by Math-X (First understand the problem!!!)

https://youtu.be/nKTOYne7E6Y

~Math-X

Video Solution 3 by OmegaLearn.org

https://youtu.be/UJ3esPnlI5M

Video Solution 4 by SpreadTheMathLove

https://www.youtube.com/watch?v=Svibu3nKB7E

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=V-xN8Njd_Lc

~NiuniuMaths

Video Solution by CosineMethod [🔥Fast and Easy🔥]

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

See Also

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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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