Difference between revisions of "2024 AMC 8 Problems/Problem 17"
Hsnacademy (talk | contribs) (→Solution 2) |
Nimitjalan (talk | contribs) m (→Solution 1) |
||
Line 26: | Line 26: | ||
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> | + | 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>. |
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. |
Latest revision as of 11:55, 12 September 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Video Solution (A Clever Explanation You’ll Get Instantly)
- 5 Video Solution 1 by Math-X (First understand the problem!!!)
- 6 Video Solution 2 (super clear!) by Power Solve
- 7 Video Solution 3 by OmegaLearn.org
- 8 Video Solution 4 by SpreadTheMathLove
- 9 Video Solution by NiuniuMaths (Easy to understand!)
- 10 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 11 Video Solution by Interstigation
- 12 See Also
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 other words, not right next to each other). In how many ways can this be done?
~Diagram by Andrei.martynau
Solution 1
If you place a king in any of the corners, they will have
spots to go and there are
corners, so
.
If you place a king in any of the
edges, they will have
spots to go and there are
edges so,
.
That gives us
spots to go into totally.
So
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
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 (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
Video Solution 3 by OmegaLearn.org
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
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.