a = 111111...............81 times

 

a =  (1/9)9999999.............81 times

 

a = (10^81 - 1)/ 9

 

a = (10^81 -1)/ 9

a = (10^27x3 - 1³)/9

a = (10²⁷ - 1)(10⁵⁴ + 10²⁷ + 1)/9

a = (10⁹ - 1)(10¹⁸ + 10⁹ + 1)(10⁵⁴ + 10²⁷ + 1) / 9

a = 9 * 111111111*(10¹⁸ + 10⁹ + 1)(10⁵⁴ + 10²⁷ + 1) / 9

a = 111111111*(10¹⁸ + 10⁹ + 1)(10⁵⁴ + 10²⁷ + 1) .......(1)

now any no. is divisible by 3 if sum of the digits is 3 or integral multiple of 3.

Further for divisibility by 9 the sum of digits should be 9.

Thus, 111111111 = 9 * m = 3²(m)  ........(2)

10¹⁸ + 10⁹ + 1 = 3*n      ( However, this no.is not divisible by 9)  .....(3)

10⁵⁴ + 10²⁷ + 1 = 3*p    ( However, this no.is not divisible by 9)   .....(4)

Where m,n and p are +ve integers not divisible by 3

Combining (1), (2), (3) and (4) we obtain

a = 3⁴ (mnp) = 81mnp

so a is divisible by 81 but not by 243 = 3⁵

so option (3) is correct