(64-(8*3))*7 = 280
(64-(8*3)-2)*5 = 190
(64-(8*3)-4)*3 = 108
(64-(8*3)-6)*1 = 34
280 + 190 + 108 +34 = 612
612 * 4 = 2448?
Chess with Maths
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Mar 15, 2024
0
#23
@stockOfHey , I am thinking that you are calculating for one queen you need to calculate for two queen .
Do you mean a queen cannot capture the other?
Sorry, I am afraid that is my answer...
Are the number of squares controlled by a queen in different file remain the same?
Are the number of squares controlled by a queen in different rank remain the same?
Are the number of squares controlled by a queen in each diagonal on each position the same?
If the answer on all of these questions are yes, why? If not, why?
Based on observation... the squares of a file controlled by a queen remain the same...
the squares of a rank controlled by a queen remain the same...
the squares of a diagonal controlled by a queen increases by two as you move closer to center from the edge...
Are these things right? or not? Why?
Because I forgot that the position of the queen is repeated on calculation... I therefore conclude that the answer must have an additional of 64*(3-1) = 128
2448+128 = 2576?...
Thank You for saying that I am wrong Ashvath23... without you, I made a mistake...
Then, I forgot the other queen... Thanks for optimissed... I must correct myself again...
2576*2 = 5152?
Mar 15, 2024
0
#31
StockOfHey wrote:
Because I forgot that the position of the queen is repeated on calculation... I therefore conclude that the answer must have an additional of 64*(3-1) = 128
2448+128 = 2576?...
Thank You for saying that I am wrong Ashvath23... without you, I made a mistake...
Then, I forgot the other queen... Thanks for optimissed... I must correct myself again...
2576*2 = 5152?
Yes, that's correct answer . 
TY!... It is hard to be duhmb... Thanks that there are people like you that I can rely on...
Mar 15, 2024
0
#35
Mr_Mathematician wrote:
Wind wrote:
I've always tried to do this on a chessboard but never could!
At max I could put 7 queens not attacking each other, 8 is very hard.
Yet, very surprised that there are 42 ways 🤯
I think you misunderstood the question , there are only two queens on board
Oops! Indeed, I thought it was 8 queens.
Still surprised that there are only 42 ways, very nice!
Mar 15, 2024
0
#36
Optimissed wrote:
Wind wrote:
Mr_Mathematician wrote:
Wind wrote:
I've always tried to do this on a chessboard but never could!
At max I could put 7 queens not attacking each other, 8 is very hard.
Yet, very surprised that there are 42 ways 🤯
I think you misunderstood the question , there are only two queens on board
Oops! Indeed, I thought it was 8 queens.
Still surprised that there are only 42 ways, very nice!
No, you misunderstood in more ways than one . The answer is 5152 ways.
OMG that's just fantastic!! 🤯
Chess and Math, what a fine mixture.
Mar 15, 2024
0
#37
Optimissed wrote:
I spent 5 months in India when I was 24 years old, which is very nearly 50 years ago. I spent my 25th birthday climbing a snow peak in thick mist. It was only a small one, about 16000 feet.
Anyway, in that time I learned to understand some of the ways that Indian people express themselves. The way that question was phrased was probably difficult for many Westerners to understand. The idea was "how many different ways can you place a white queen and a black queen on a board in such a way that the two queens do not attack each other?" So you have to work it out for one of the queens and then you double it to cover both queens. The closer you place one queen towards the centre of the board, the less positions there are where you can place the other queen and it isn't attacked. In fact, for each square nearer the centre that you place the first queen, the number of positions you can place the second queen decreases by two, showing that a queen becomes more powerful, the nearer it gets to the centre of the board.
I didn't know any of this before I saw this question yesterday but I worked it all out in about ten minutes.
Yes! The same feel !
Placing 8 queens is also possible with 2 knight zone at tow side and adjust 2 queens in middle accordingly
This make 8 queens on board with no problem
There can be more than one way in it
Many people might think what is use of that but in reality software engineer need such calculations
After some thought... I made a mistake...