(In reply to

Solution by K Sengupta)

Alternatively, expanding by Binomial Theorem, we have:

2^95

= (33-1)^19

= (33^19 – comb(19, 1)*33^18 +......- comb(19,17)*33^2 + comb(19,18)*33 – 1)

= M(99) + 19*33 -1,

= M(99) + 18*33 + (33-1)

= M(99) + 32

Accordingly,

2^95 (mod 99) = 32, so that:

2^99 (mod 99) = (32*16) mod 99

But, 32*16

= (33-1)(15+1)

= 33*15 +(33-15-1)

= M(99) + 17, since 15 is divisible by 3, so that 33*15 is divisible by 99

Accordingly,

2^99 (mod 99) = 17

Consequently, the required remainder is 17.