Is there an algorithm for calculating the multiplicative order of x modulo y (for y < 1000) that doesn't require a BigInteger type?

Question

The algorithm I'm using at the moment runs into extremely high numbers very quickly. A step in the algorithm I'm to raises x to the result of the totient function applied to y. The result is that you can run into very large numbers.

Eg. When calculating the multiplicative order of 10 modulo 53:

10^totient(53) == 10^52 == 1 * 10^52

The following algorithm fares a bit better either in terms of avoiding large numbers, but it still fails where 10^mOrder is greater than the capacity of the data type:

  mOrder = 1
  while 10^mOrder % 53 != 1
      if mOrder >= i
          mOrder = 0;
          break
      else
          mOrder = mOrder + 1

Answer 1

Using Modular exponentiation, it is possible to calculate (10 ^ mOrder % 53) or in general, any (a ^ b mod c) without getting values much bigger than c. See Wikipedia for details, there's this sample code, too:

Bignum modpow(Bignum base, Bignum exponent, Bignum modulus) {

    Bignum result = 1;

    while (exponent > 0) {
        if ((exponent & 1) == 1) {
            // multiply in this bit's contribution while using modulus to keep result small
            result = (result * base) % modulus;
        }
        // move to the next bit of the exponent, square (and mod) the base accordingly
        exponent >>= 1;
        base = (base * base) % modulus;
    }

    return result;
}

Answer 2

Why exponentiate? Can't you just multiply modulo n in a loop?

(defun multiplicative-order (a n)
  (if (> (gcd a n) 1)
      0
      (do ((order 1 (+ order 1))
           (mod-exp a (mod (* mod-exp a) n)))
          ((= mod-exp 1) order))))

Or, in ptheudo (sic) code:

def multiplicative_order (a, n) :
    if gcd (a, n) > 1 :
        return 0
      else:
        order = 1
        mod_exp = a
        while mod_exp != 1 :
            order += 1
            mod_exp = (mod_exp * a) mod n
        return order