At the outset, we note that 33^2 is divisible by 99, so that 33^x is divisible by 99 for all positive integers x >= 2.

Now,

2^15 = 32^3

=(33-1)^3

= (33^3 – 33*3(33 -1) – 1)

= M(99) – 1

Thus: 2^15 = -1 (mod 99)

or, 2^90 = 1 (mod 99)

Again:

2^9

= 16*32

= (15+1)(33-1)(mod 99)

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

= M(99) + 17, and so: 2^9 = 17 (mod 99); since 15 is divisible by 3, so that 33*15 is divisible by 99

Thus, 2^99

= (2^90)*(2^9) (mod 99)

= 1*17 (mod 99)

= 17

Consequently, the required remainder is 17.

*Edited on ***August 14, 2008, 2:04 am**