Solution:
2^133 = 2^(7*19) = (2^7)^19 = (128)^19 = (133 - 5)^19.
Now (133 - 5)^19 = 133^19 - 19C1*(133^18)*(5^1) + 19C2*(133^17)*(5^2) .............+19C18*(133^1)*(5^18) - 5^19.

In the above binomial expansion all the terms except -5^19 is divisible by 133.
So what we need to do is to find the remainder when 5^19 is divided by 133 and subtract that from 133.

Proceed to find the remainder when 5^19 is divided by 133.
Now 5^19 = (5^18)*5 = (125^6)*5 = {(133 - 8)^6}*5.
Now on expanding {(133 - 8)^6}*5 all other terms are divisible by 133 except for the last term which is (8^6)*5 = {(512)^2}*5 = {(532 - 20)^2}*5.
Now on expanding {(532 - 20)^2}*5 , since 532 is 133*4, all other terms are divisible by 133 except for last term which is (20^2)*5 = 2000.
On dividing 2000 by 133, the remainder is 5.
On subtracting 5 from 133, we have 133 - 5 = 128 as remainder.