On an \(n \times n\) chessboard, the total number of rectangles which are not squares is 350 . Then, the…

On an \(n \times n\) chessboard, the total number of rectangles which are not squares is 350 . Then, the number of white squares on the chessboard is .......
  1. 32
  2. 50
  3. 18
  4. 72

Solution

Number of rectangles in a chess board (including squares) \(=\Sigma n^3\) \(=\frac{n^2(n+1)^2}{4}\) Number of squares in a chess board \(=\Sigma n^2=\frac{n(n+1)(2 n+1)}{6}\) So, Number of rectangles which are not squares \(\begin{aligned} &=\Sigma n^3-\Sigma n^2 \\ &=\frac{n^2(n+1)^2}{4}-n \frac{(n+1)(2 n+1)}{6}=350 \\ & \Rightarrow \quad n\left(\frac{n+1}{2}\right)\left(\frac{n(n+1)}{2}-\frac{2 n+1}{3}\right)=350 \end{aligned}\) \(\begin{aligned} \Rightarrow & & n\left(\frac{n+1}{2}\right)\left(\frac{3 n^2+3 n-4 n-2}{6}\right) & =350 \\ \Rightarrow & & n(n+1)\left(3 n^2-n-2\right) & =350 \times 6 \times 2 \\ \Rightarrow & & n(n+1)\left(3 n^2-3 n+2 n-2\right) & =350 \times 6 \times 2 \\ \Rightarrow & & n(n+1)(3 n(n-1)+2(n-1) & =350 \times 6 \times 2 \\ \Rightarrow & & n(n+1)(n-1)(3 n+2) & =350 \times 6 \times 2 \\ \Rightarrow & & (n-1) n(n+1)(3 n+2) & =5 \times 6 \times 7 \times 20 \end{aligned}\) By comparison we get \(\therefore \quad n=6\) Hence, total number of black and white squares \(=n \times n=6 \times 6=36\) Half of there are \(\text {white }=\frac{36}{2}=18 \text { squares. }\)

Asked in: AP EAMCET 2020 (17 Sep Shift 2)

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