(4 + 5 + 2) - (1 + 6 + 3) = 1, not divisible by 11.

(2 + 6 + 4) - (4 + 5 + 2) = 1, not divisible by 11.

(4 + 6 + 1) - (2 + 5 + 3) = 1, not divisible by 11.

(4 + 6 + 1) - (2 + 5 + 4) = 0, So, 415624 is divisible by 11.

19657         Let x - 53651  = 9999

 33994         Then, x = 9999 + 53651 = 63650

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 53651

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Let S_n =(1 + 2 + 3 + ... + 45). This is an A.P. in which a =1, d =1, n = 45.

S_n =  \(\frac{n}{2}\left[2a+(n-1)d\right]=\frac{45}{2}\times\left[2\times1+(45-1)\times1\right]=\left(\frac{45}{2}\times46\right)=45\times23\)

= 45 x (20 + 3)

= 45 x 20 + 45 x 3

= 900 + 135

=1035

Shorcut Method:

S_n = \(\frac{n(n+1)}{2} = \frac{45(45+1)}{2}=1035\)

24 = 3 x8, where 3 and 8 co-prime.

Clearly, 35718 is not divisible by 8, as 718 is not divisible by 8.

Similarly, 63810 is not divisible by 8 and 537804 is not divisible by 8.

Consider option (D),

Sum of digits = (3 + 1 + 2 + 5 + 7 + 3 + 6) = 27, which is divisible by 3.

Also, 736 is divisible by 8.

 3125736 is divisible by (3 x 8), i.e., 24.

Given Exp. = \(\frac{(a^{2}+b^{2}-ab)}{(a^{3}+b^{3})}=\frac{1}{(a+b)}=\frac{1}{(753+247)}=\frac{1}{1000}\)

x + 3699 + 1985 - 2047 = 31111

   x + 3699 + 1985 = 31111 + 2047

   x + 5684 = 33158

   x = 33158 - 5684 = 27474.

Sum of digits = (4 + 8 + 1 + x + 6 + 7 + 3) = (29 + x), which must be divisible by 9.

   x = 7.

(Local value of 7) - (Face value of 7) = (70000 - 7) = 69993

Let the required fraction be x. Then \(\frac{1}{x}-x = \frac{9}{20}\)

Therefore \(\frac{1-x^{2}}{x} = \frac{9}{20}\)

   20 - 20x² = 9x

      20x² + 9x - 20 = 0

      20x² + 25x - 16x - 20 = 0

      5x(4x + 5) - 4(4x + 5) = 0

(4x + 5)(5x - 4) = 0

\(x = \frac{4}{5}\)