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== | ||
+ | |||
+ | 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 <math>\binom{4}{2}=6</math> | ||
+ | |||
+ | Corner-edge: For each corner, there are two edges that don't border it, <math>4\cdot2=8</math> | ||
+ | |||
+ | Edge-edge: The only possible combinations of this that work are top-bottom and left-right edges, so <math>2</math> for this type | ||
+ | |||
+ | |||
+ | <math>6+8+2=16</math> | ||
+ | |||
+ | 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 17:28, 26 January 2024
Contents
[hide]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
Multiply by two to account for arrangements of colors to get
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 by Math-X (First understand the problem!!!)
~Math-X