How to calculate modulus of large numbers?

Question

How to calculate modulus of 5^55 modulus 221 without much use of calculator?

I guess there are some simple principles in number theory in cryptography to calculate such things.

Answer 1

Okay, so you want to calculate a^b mod m. First we'll take a naive approach and then see how we can refine it.

First, reduce a mod m. That means, find a number a1 so that 0 <= a1 < m and a = a1 mod m. Then repeatedly in a loop multiply by a1 and reduce again mod m. Thus, in pseudocode:

a1 = a reduced mod m
p = 1
for(int i = 1; i <= b; i++) {
    p *= a1
    p = p reduced mod m
}

By doing this, we avoid numbers larger than m^2. This is the key. The reason we avoid numbers larger than m^2 is because at every step 0 <= p < m and 0 <= a1 < m.

As an example, let's compute 5^55 mod 221. First, 5 is already reduced mod 221.

1. 1 * 5 = 5 mod 221
2. 5 * 5 = 25 mod 221
3. 25 * 5 = 125 mod 221
4. 125 * 5 = 183 mod 221
5. 183 * 5 = 31 mod 221
6. 31 * 5 = 155 mod 221
7. 155 * 5 = 112 mod 221
8. 112 * 5 = 118 mod 221
9. 118 * 5 = 148 mod 221
10. 148 * 5 = 77 mod 221
11. 77 * 5 = 164 mod 221
12. 164 * 5 = 157 mod 221
13. 157 * 5 = 122 mod 221
14. 122 * 5 = 168 mod 221
15. 168 * 5 = 177 mod 221
16. 177 * 5 = 1 mod 221
17. 1 * 5 = 5 mod 221
18. 5 * 5 = 25 mod 221
19. 25 * 5 = 125 mod 221
20. 125 * 5 = 183 mod 221
21. 183 * 5 = 31 mod 221
22. 31 * 5 = 155 mod 221
23. 155 * 5 = 112 mod 221
24. 112 * 5 = 118 mod 221
25. 118 * 5 = 148 mod 221
26. 148 * 5 = 77 mod 221
27. 77 * 5 = 164 mod 221
28. 164 * 5 = 157 mod 221
29. 157 * 5 = 122 mod 221
30. 122 * 5 = 168 mod 221
31. 168 * 5 = 177 mod 221
32. 177 * 5 = 1 mod 221
33. 1 * 5 = 5 mod 221
34. 5 * 5 = 25 mod 221
35. 25 * 5 = 125 mod 221
36. 125 * 5 = 183 mod 221
37. 183 * 5 = 31 mod 221
38. 31 * 5 = 155 mod 221
39. 155 * 5 = 112 mod 221
40. 112 * 5 = 118 mod 221
41. 118 * 5 = 148 mod 221
42. 148 * 5 = 77 mod 221
43. 77 * 5 = 164 mod 221
44. 164 * 5 = 157 mod 221
45. 157 * 5 = 122 mod 221
46. 122 * 5 = 168 mod 221
47. 168 * 5 = 177 mod 221
48. 177 * 5 = 1 mod 221
49. 1 * 5 = 5 mod 221
50. 5 * 5 = 25 mod 221
51. 25 * 5 = 125 mod 221
52. 125 * 5 = 183 mod 221
53. 183 * 5 = 31 mod 221
54. 31 * 5 = 155 mod 221
55. 155 * 5 = 112 mod 221

Therefore, 5^55 = 112 mod 221.

Now, we can improve this by using exponentiation by squaring; this is the famous trick wherein we reduce exponentiation to requiring only log b multiplications instead of b. Note that with the algorithm that I described above, the exponentiation by squaring improvement, you end up with the right-to-left binary method.

a1 = a reduced mod m
p = 1
while (b > 0) {
     if (b is odd) {
         p *= a1
         p = p reduced mod m
     }
     b /= 2
     a1 = (a1 * a1) reduced mod m
}

Thus, since 55 = 110111 in binary

1. 1 * (5^1 mod 221) = 5 mod 221
2. 5 * (5^2 mod 221) = 125 mod 221
3. 125 * (5^4 mod 221) = 112 mod 221
4. 112 * (5^16 mod 221) = 112 mod 221
5. 112 * (5^32 mod 221) = 112 mod 221

Therefore the answer is 5^55 = 112 mod 221. The reason this works is because

55 = 1 + 2 + 4 + 16 + 32

so that

5^55 = 5^(1 + 2 + 4 + 16 + 32) mod 221
     = 5^1 * 5^2 * 5^4 * 5^16 * 5^32 mod 221
     = 5 * 25 * 183 * 1 * 1 mod 221
     = 22875 mod 221
     = 112 mod 221

In the step where we calculate 5^1 mod 221, 5^2 mod 221, etc. we note that 5^(2^k) = 5^(2^(k-1)) * 5^(2^(k-1)) because 2^k = 2^(k-1) + 2^(k-1) so that we can first compute 5^1 and reduce mod 221, then square this and reduce mod 221 to obtain 5^2 mod 221, etc.

The above algorithm formalizes this idea.

Answer 2

To add to Jason's answer:

You can speed the process up (which might be helpful for very large exponents) using the binary expansion of the exponent. First calculate 5, 5^2, 5^4, 5^8 mod 221 - you do this by repeated squaring:

 5^1 = 5(mod 221)
 5^2 = 5^2 (mod 221) = 25(mod 221)
 5^4 = (5^2)^2 = 25^2(mod 221) = 625 (mod 221) = 183(mod221)
 5^8 = (5^4)^2 = 183^2(mod 221) = 33489 (mod 221) = 118(mod 221)
5^16 = (5^8)^2 = 118^2(mod 221) = 13924 (mod 221) = 1(mod 221)
5^32 = (5^16)^2 = 1^2(mod 221) = 1(mod 221)

Now we can write

55 = 1 + 2 + 4 + 16 + 32

so 5^55 = 5^1 * 5^2 * 5^4 * 5^16 * 5^32 
        = 5   * 25  * 625 * 1    * 1 (mod 221)
        = 125 * 625 (mod 221)
        = 125 * 183 (mod 183) - because 625 = 183 (mod 221)
        = 22875 ( mod 221)
        = 112 (mod 221)

You can see how for very large exponents this will be much faster (I believe it's log as opposed to linear in b, but not certain.)

Answer 3

/* The algorithm is from the book "Discrete Mathematics and Its
   Applications 5th Edition" by Kenneth H. Rosen.
   (base^exp)%mod
*/

int modular(int base, unsigned int exp, unsigned int mod)
{
    int x = 1;
    int power = base % mod;

    for (int i = 0; i < sizeof(int) * 8; i++) {
        int least_sig_bit = 0x00000001 & (exp >> i);
        if (least_sig_bit)
            x = (x * power) % mod;
        power = (power * power) % mod;
    }

    return x;
}

Answer 4

5^55 mod221

= (   5^10         * 5^10         * 5^10         * 5^10          * 5^10          * 5^5) mod221    

= ( ( 5^10) mod221 * 5^10         * 5^10         * 5^10          * 5^10          * 5^5) mod221 

= (   77           * 5^10         * 5^10         * 5^10          * 5^10          * 5^5) mod221   

= ( ( 77           * 5^10) mod221 * 5^10         * 5^10          * 5^10          * 5^5) mod221 

= (   183                         * 5^10         * 5^10          * 5^10          * 5^5) mod221 

= ( ( 183                         * 5^10) mod221 * 5^10          * 5^10          * 5^5) mod221 

= (   168                                        * 5^10          * 5^10          * 5^5) mod221 

= ( ( 168                                        * 5^10) mod 221 * 5^10          * 5^5) mod221 

= (   118                                                        * 5^10          * 5^5) mod221 

= ( ( 118                                                        * 5^10) mod 221 * 5^5) mod221 

= (   25                                                                         * 5^5) mod221 

=     112

Answer 5

Chinese Remainder Theorem comes to mind as an initial point as 221 = 13 * 17. So, break this down into 2 parts that get combined in the end, one for mod 13 and one for mod 17. Second, I believe there is some proof of a^(p-1) = 1 mod p for all non zero a which also helps reduce your problem as 5^55 becomes 5^3 for the mod 13 case as 13*4=52. If you look under the subject of "Finite Fields" you may find some good results on how to solve this.

EDIT: The reason I mention the factors is that this creates a way to factor zero into non-zero elements as if you tried something like 13^2 * 17^4 mod 221, the answer is zero since 13*17=221. A lot of large numbers aren't going to be prime, though there are ways to find large primes as they are used a lot in cryptography and other areas within Mathematics.

Answer 6

What you're looking for is modular exponentiation, specifically modular binary exponentiation. This wikipedia link has pseudocode.

Answer 7

This is part of code I made for IBAN validation. Feel free to use.

    static void Main(string[] args)
    {
        int modulo = 97;
        string input = Reverse("100020778788920323232343433");
        int result = 0;
        int lastRowValue = 1;

        for (int i = 0; i < input.Length; i++)
        {
            // Calculating the modulus of a large number Wikipedia http://en.wikipedia.org/wiki/International_Bank_Account_Number                                                                        
            if (i > 0)
            {
                lastRowValue = ModuloByDigits(lastRowValue, modulo);
            }
            result += lastRowValue * int.Parse(input[i].ToString());
        }
        result = result % modulo;
        Console.WriteLine(string.Format("Result: {0}", result));            
    }

    public static int ModuloByDigits(int previousValue, int modulo)
    {
        // Calculating the modulus of a large number Wikipedia http://en.wikipedia.org/wiki/International_Bank_Account_Number                        
        return ((previousValue * 10) % modulo);
    }
    public static string Reverse(string input)
    {
        char[] arr = input.ToCharArray();
        Array.Reverse(arr);
        return new string(arr);
    }

Answer 8

Jason's answer in Java (note i < exp).

private static void testModulus() {
    int bse = 5, exp = 55, mod = 221;

    int a1 = bse % mod;
    int p = 1;

    System.out.println("1. " + (p % mod) + " * " + bse + " = " + (p % mod) * bse + " mod " + mod);

    for (int i = 1; i < exp; i++) {
        p *= a1;
        System.out.println((i + 1) + ". " + (p % mod) + " * " + bse + " = " + ((p % mod) * bse) % mod + " mod " + mod);
        p = (p % mod);
    }

}

Answer 9

Just provide another implementation of Jason's answer by C.

After discussing with my classmates, based on Jason's explanation, I like the recursive version more if you don't care about the performance very much:

For example:

#include<stdio.h>

int mypow( int base, int pow, int mod ){
    if( pow == 0 ) return 1;
    if( pow % 2 == 0 ){
        int tmp = mypow( base, pow >> 1, mod );
        return tmp * tmp % mod;
    }
    else{
        return base * mypow( base, pow - 1, mod ) % mod;
    }
}

int main(){
    printf("%d", mypow(5,55,221));
    return 0;
}

Answer 10

This is called modular exponentiation(https://en.wikipedia.org/wiki/Modular_exponentiation).

Let's assume you have the following expression:

19 ^ 3 mod 7

Instead of powering 19 directly you can do the following:

(((19 mod 7) * 19) mod 7) * 19) mod 7

But this can take also a long time due to a lot of sequential multipliations and so you can multiply on squared values:

x mod N -> x ^ 2 mod N -> x ^ 4 mod -> ... x ^ 2 |log y| mod N

Modular exponentiation algorithm makes assumptions that:

x ^ y == (x ^ |y/2|) ^ 2 if y is even
x ^ y == x * ((x ^ |y/2|) ^ 2) if y is odd

And so recursive modular exponentiation algorithm will look like this in java:

/**
* Modular exponentiation algorithm
* @param x Assumption: x >= 0
* @param y Assumption: y >= 0
* @param N Assumption: N > 0
* @return x ^ y mod N
*/
public static long modExp(long x, long y, long N) {
    if(y == 0)
        return 1 % N;

    long z = modExp(x, Math.abs(y/2), N);

    if(y % 2 == 0)
        return (long) ((Math.pow(z, 2)) % N);
    return (long) ((x * Math.pow(z, 2)) % N);
}

Special thanks to @chux for found mistake with incorrect return value in case of y and 0 comparison.