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
Solution
