For example I want to calculate (reasonably efficiently)
2^1000003 mod 12321
And finally I want to do (2^1000003 - 3) mod 12321. Is there any feasible way to do this?
Basic modulo properties tell us that
1) a + b (mod n)
is (a (mod n)) + (b (mod n)) (mod n)
, so you can split the operation in two steps
2) a * b (mod n)
is (a (mod n)) * (b (mod n)) (mod n)
, so you can use modulo exponentiation (pseudocode):
x = 1
for (10000003 times) {
x = (x * 2) % 12321; # x will never grow beyond 12320
}
Of course, you shouldn't do 10000003 iterations, just remember that 21000003 = 2 * 21000002 , and 21000002 = (2500001)2 and so on...