6n² + 6n = 6n(n + 1)
Hence 6n² + 6n is always divisible by 6 and 12 (∵ remember that n(n + 1) is always even)

(a+b)2+(a−b)2=2(a2+b2)

1092+912=(100+9)2+(100−9)2=2(1002+92)=2(10000+81)=20162

(xn+1) is divisible by (x + 1) only when n is odd

=> (67⁶⁷ + 1) is divisible by (67 + 1)

=> (67⁶⁷ + 1) is divisible by 68

=> (67⁶⁷ + 1) ÷  68 gives a remainder of 0

=> [(67⁶⁷ + 1) + 66] ÷ 68 gives a remainder of 66

\(\frac{(912+643)^{2}+(912-643)^{2}}{(912\times912+643\times643)} = \frac{(912+643)^{2}+(912-643)^{2}}{(912^{2}+643^{2})}= \frac{2(912^{2}+643^{2})}{(912^{2}+643^{2})}=2\)

23341379 x 72 = 23341379(70 + 2) = (23341379 x 70) + (23341379 x 2)
= 1633896530 + 46682758 = 1680579288

Replacing * by x
5 x 2 is divisible by 2 (Reference : Divisibility by 2 rule)
For 5 x 2 to be divisible by 3, 5 + x + 2 shall be divisible by 3 (Reference : Divisibility by 3 rule)
=> 7 + x shall be divisible by 3
=> x can be 2 or 5 or 8
From the given choices, answer = 2

\(\left(1-\frac{1}{n}\right)+\left(1-\frac{2}{n}\right)+\left(1-\frac{3}{n}\right)+... up to n terms \) 

= \(\left(1+1+1+... up to rerms \right) - \left(\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+...up to rerms\right)\)

= \(n-\frac{1}{n}\left(1+2+3+...up to rerms\right)\)

= \(n-\frac{1}{n}\left[\frac{n(n+1)}{2}\right]\)

= \(n-\frac{(n+1)}{2}\)

= \(\frac{(2n-n-1)}{2}\)   

= \(\frac{n-1}{2}\)

Let x + 3699 + 1985 - 2047 = 31111
=> x = 31111 - 3699 - 1985 + 2047 = 27474