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

(Video Solution 3 by OmegaLearn.org)
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(A) <math>20</math>  (B) <math>24</math>  (C) <math>27</math>  (D) <math>28</math>  (E) <math>32</math>
 
(A) <math>20</math>  (B) <math>24</math>  (C) <math>27</math>  (D) <math>28</math>  (E) <math>32</math>
  
==Solution 1== by Math 645
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==Solution 1==
  
Corners have 5 spots to go and 4 corners so 5*4=20.
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Corners have <math>5</math> spots to go and <math>4</math> corners so <math>5 \times 4=20</math>.
Sides have 3 spots to go and 4 sides so 3*4=12
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Sides have <math>3</math> spots to go and <math>4</math> sides so <math>3 \times 4=12</math>
20+12=32 in total.
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<math>20+12=32</math> in total.
E (32) is the answer.
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<math>\boxed{\textbf{(E)} 32)}</math> is the answer.
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~andliu766
  
 
==Video Solution 1 by Math-X (First understand the problem!!!)==
 
==Video Solution 1 by Math-X (First understand the problem!!!)==

Revision as of 18:19, 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

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