Fool's problem of the day, A tiled floor has dimensions \(m \times m \) sq.m. The dimensions of the tile used are \(n\times n\) sq.m. All the tiles used are green tiles except the diagonals tiles which are red. After some years some green tiles are replaced by red tiles to form an alternate red and green tile pattern.If \(m \neq n \) and total number of tiles are odd then how many green tiles are removed?

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Fool's problem of the day, A tiled floor has dimensions \(m \times m \) sq.m. The dimensions of the tile used are \(n\times n\) sq.m. All the tiles used are green tiles except the diagonals tiles which are red. After some years some green tiles are replaced by red tiles to form an alternate red and green tile pattern.If \(m \neq n \) and total number of tiles are odd then how many green tiles are removed?

Mathematics
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PS: http://finance.groups.yahoo.com/group/ascent4cat/message/2900 http://finance.groups.yahoo.com/group/ascent4cat/message/2915 These solutions are incorrect.
I am not so sure. is it \(\large m(\frac{m}{n*2} - \frac{3}{2})\)
I am sorry Ishaan :(

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Umm m is the dimension of the tiled floor, n the dimension of a tile. So, total number of tiles in a column or row (since it's a square) must be \(\frac{m}{n}\). Hmm then we have boundary of red tiles and we wish to create an alternate pattern of Red and Green (Horrible color combination btw). The number of red tiles would be always 1 greater than Green ones after the completion of alternate pattern.(initially we have two tiles in red of each row or column) So, in one row \(\large\frac{\frac{m}{n} -2 -1}{2}\) must be green. We have m-2 column not m sorry. Hence, the answer should be \(\large \frac{\frac{m}{n} - 3}{2} (m-2)\) Am I doing it right?
Oh Wait.. I did something again umm It should be \[\left(\frac{\frac{m}{n} -1}{2}\right) ( m-2)\]
Oh no I did something again \[\frac{\frac{m}{n} - 1}{2}, \text{ is the number of green tiles} \implies \text{number of red tiles must be } \frac{m}{n}- 2 - \frac{m}{2n} - \frac{1}{2} \] \[\frac{m}{n}- 2 - \frac{m}{2n} - \frac{1}{2} = \frac{m}{2n} - \frac{3}{2}\] So number of tiles to be replaced are \(\left( \frac{m}{2n} - \frac{3}{2} \right) (m-2)\)
I can't believe I took so much time to solve a CAT question and still don't know if it's right
Sorry again it's not what I got, I am getting confused again and again \[\left( \frac{m}{2n} - \frac{3}{2} \right) (\frac{m}{n}-2)\] Because m/n is the number of rows and column not m
Doesn't seem correct, if you want I can give you the answer.
Impossibru I checked my logic for even number of rows and column
Hmm lemme check again
For Odd this time
Mere working for small m,n doesn't mean a conclusive proof.
The correct answer is \( \frac{1}{2} n^2 \left((m-2 n)^2-n^2\right)\)
\[\left( \frac{m}{2n} - \frac{1}{2} \right) (\frac{m}{n}-2)\] Umm but you don't have n in the denominator? How do you get rows and column?
\[\frac{1}{2}* \frac{1}{n^2} (m-n)(m-2n) = \frac{1}{2n^2} ( m^2 - 2mn -nm + 2n^2)\]This is all I am getting umm looks like I will have to check again
\[\left( \frac{m}{2n} - \frac{3}{2} \right) (\frac{m}{n}-2)\] \[\frac{1}{2}\left( \frac{m}{n} - \frac{3}{1} \right) (\frac{m}{n}-2)=\frac{1}{2}* \frac{1}{n^2} (m-3n)(m-2n) = \frac{1}{2n^2} ( m^2 - 2mn -3nm + 6n^2)\] Lets take the case of a floor where m = 99 and n = 1, total rows = columns = 99. 97 tiles in each row are green, 2 are red. We need an alternate pattern which consists of 50 red tiles and 49 green. Since 2 were already red 48 tiles were replaced in each row and there are 99-2 columns on the floor (first and last columns are completely red so no replacement). total tiles to be replaced = 48* (99-2) = 48*97 By my formula I get the same \[\frac{1}{2}(99-3)(99 -2) = 48*97=4656\] And with yours (foolformath's solution) \(\frac{1}{2} n^2 \left((m-2 n)^2-n^2\right)\), I am getting \[\frac{1}{2}* 1* ((99-2)^2 - 1) = 4704\] What did I miss? Please tell me where I am wrong?
Oh I get it Diagnol's tiles are red, I thought the floor was bounded by red tiles. Silly me I should learn to read questions properly
\[f(tiles)= \left\{\begin{array}{r}\left( \frac{1}{2}\right)\left( \frac{m}{2n} - 3 \right) \left(\frac{m}{n}-2\right) + \frac{m}{2n} -\frac{3}{2}, \text{if} &\frac{m}{n} \in \{(4k-3), k \in \mathbb{N}\} \\ \left(\frac{1}{2}\right)\left( \frac{m}{2n} - 3 \right) \left(\frac{m}{n}-2\right) + \frac{m}{2n} - \frac{1}{2}, \text{if}& \frac{m}{n} \in \{(4k -1), k \in \mathbb{N}\} \end{array} \right.\] \(f(tiles)\) is the function for number of tiles lol I know I have made it so complicated but it works! :-D lol
I can't stop laughing lol I made it so bad and yet it works.
Nah it doesn't works I committed some error again I will check out this problem later
work*
\[f(tiles)= \left\{\begin{array}{r}\left( \frac{1}{2}\right)\left( \frac{m}{n} - 3 \right) \left(\frac{m}{n}-2\right) + \frac{m}{2n} -\frac{3}{2}, \text{if} &\frac{m}{n} \in \{(4k-3), k \in \mathbb{N}\} \\ \left(\frac{1}{2}\right)\left( \frac{m}{n} - 3 \right) \left(\frac{m}{n}-2\right) + \frac{m}{2n} - \frac{1}{2}, \text{if}& \frac{m}{n} \in \{(4k -1), k \in \mathbb{N}\} \end{array} \right.\] Correction* Note. I am testing it
I have it FINALLY! \[\frac{1}{2} \left(\left(\frac{m}{n}-2\right)\left(\frac{m}{n}-3\right) + \frac{m}{n} -3\right) = \frac{1}{2} \left(\frac{m}{n} -3\right) \left( \frac{m}{n} -1\right)\]
Yay! I got it! The closed form! Yay!
Took me time, pathetic :-/
Hehe, should i post my solution?
Yeah, you should
Okay, after few hours :)
Done! :-)

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