Difference between revisions of "2024 AMC 8 Problems/Problem 17"
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Sides have <math>3</math> spots to go and <math>4</math> sides so <math>3 \times 4=12</math> | Sides have <math>3</math> spots to go and <math>4</math> sides so <math>3 \times 4=12</math> | ||
<math>20+12=32</math> in total. | <math>20+12=32</math> in total. | ||
| − | <math>\boxed{\textbf{(E)} 32 | + | <math>\boxed{\textbf{(E)} 32}</math> is the answer. |
~andliu766 | ~andliu766 | ||
| Line 44: | Line 44: | ||
==Video Solution 4 by SpreadTheMathLove== | ==Video Solution 4 by SpreadTheMathLove== | ||
https://www.youtube.com/watch?v=Svibu3nKB7E | https://www.youtube.com/watch?v=Svibu3nKB7E | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2024|num-b=16|num-a=18}} | ||
| + | {{MAA Notice}} | ||
Revision as of 20:01, 26 January 2024
Contents
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 
x grid attacks all 
other squares, as shown below. Suppose a white king and a black king are placed on different squares of a x grid so that they do not attack each other. In how many ways can this be done?
(A) 
(B) 
(C) 
(D) 
(E) 
Solution 1
Corners have 
spots to go and 
corners so 
.
Sides have spots to go and sides so 
in total.
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 
Corner-edge: For each corner, there are two edges that don't border it, 
Edge-edge: The only possible combinations of this that work are top-bottom and left-right edges, so 
for this type
Multiply by two to account for arrangements of colors to get 
~ c_double_sharp
Video Solution 1 (super clear!) by Power Solve
Video Solution 2 by Math-X (First understand the problem!!!)
~Math-X
Video Solution 3 by OmegaLearn.org
Video Solution 4 by SpreadTheMathLove
https://www.youtube.com/watch?v=Svibu3nKB7E
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. 
