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

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<math>6+8+2=16</math> Multiply by two to account for arrangements of colors to get <math>E) 32</math>
 
<math>6+8+2=16</math> Multiply by two to account for arrangements of colors to get <math>E) 32</math>
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==Solution 2==
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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.
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This gives three combinations:
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Corner-corner: There are 4 corners, and none of them are touching orthogonally or diagonally, so it's <math>\binom{4}{2}=6</math>
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Corner-edge: For each corner, there are two edges that don't border it, <math>4\cdot2=8</math>
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Edge-edge: The only possible combinations of this that work are top-bottom and left-right edges, so <math>2</math> for this type
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<math>6+8+2=16</math>
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Multiply by two to account for arrangements of colors to get <math>\fbox{E) 32}</math> ~ c_double_sharp
  
 
==Video Solution 1 by Math-X (First understand the problem!!!)==
 
==Video Solution 1 by Math-X (First understand the problem!!!)==

Revision as of 18:28, 26 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?

(A) $20$ (B) $24$ (C) $27$ (D) $28$ (E) $32$

Solution 1

Corners have $5$ spots to go and $4$ corners so $5 \times 4=20$. Sides have $3$ spots to go and $4$ sides so $3 \times 4=12$ $20+12=32$ in total. $\boxed{\textbf{(E)} 32)}$ is the answer.

~andliu766

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 $2$

$6+8+2=16$ Multiply by two to account for arrangements of colors to get $E) 32$

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 by Math-X (First understand the problem!!!)

https://youtu.be/nKTOYne7E6Y

~Math-X

Video Solution 2 (super clear!) by Power Solve

https://youtu.be/SG4PRARL0TY

Video Solution 3 by OmegaLearn.org

https://youtu.be/UJ3esPnlI5M

Video Solution 4 by SpreadTheMathLove

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

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