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Difference between revisions of "2024 AMC 8 Problems/Problem 17"
m (→‎Solution 1)
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If you place a king in any of the <math>4</math> corners, they will have <math>5</math> spots to go and there are <math>4</math> corners, so <math>5 \times 4=20</math>.
 
If you place a king in any of the <math>4</math> corners, they will have <math>5</math> spots to go and there are <math>4</math> corners, so <math>5 \times 4=20</math>.
−
If you place a king in any of the <math>3</math> edges, they will have <math>3</math> spots to go and there are <math>4</math> edges so, <math>3 \times 4=12</math>.  
+
If you place a king in any of the <math>4</math> edges, they will have <math>3</math> spots to go and there are <math>4</math> edges so, <math>3 \times 4=12</math>.  
 
That gives us <math>20+12=32</math> spots to go into totally.
 
That gives us <math>20+12=32</math> spots to go into totally.
 
So <math>\boxed{\textbf{(E)} 32}</math> is the answer.
 
So <math>\boxed{\textbf{(E)} 32}</math> is the answer.

Revision as of 21:31, 28 September 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 other words, not right next to each other). In how many ways can this be done?

[asy] /* AMC8 P17 2024, revised by Teacher David */ unitsize(29pt); import math; add(grid(3,3)); pair [] a = {(0.5,0.5), (0.5, 1.5), (0.5, 2.5), (1.5, 2.5), (2.5,2.5), (2.5,1.5), (2.5,0.5), (1.5,0.5)}; for (int i=0; i<a.length; ++i) { pair x = (1.5,1.5) + 0.4*dir(225-45*i); draw(x -- a[i], arrow=EndArrow()); } label("$K$", (1.5,1.5)); [/asy]

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

~Diagram by Andrei.martynau

Solution 1

If you place a king in any of the $4$ corners, they will have $5$ spots to go and there are $4$ corners, so $5 \times 4=20$. If you place a king in any of the $4$ edges, they will have $3$ spots to go and there are $4$ edges so, $3 \times 4=12$. That gives us $20+12=32$ spots to go into totally. So $\boxed{\textbf{(E)} 32}$ is the answer. ~andliu766, captaindiamond868 (minor edits)

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 (A Clever Explanation You’ll Get Instantly)

https://youtu.be/5ZIFnqymdDQ?si=_SHs3ZXS1ZKF4-v2&t=2381

~hsnacademy

Video Solution 1 by Math-X (First understand the problem!!!)

https://youtu.be/BaE00H2SHQM?si=Q2e8OfkuzKZXmoau&t=4624

~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

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

Video Solution by Interstigation

https://youtu.be/ktzijuZtDas&t=1922

Video Solution by Dr. David

https://youtu.be/vO1rcJZzIrQ

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

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

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