You are given two integer numbers, the base a (number of digits d, such that 1 <= d <= 1000) and the index b (0 <= b <= 922*10^15). You have to find the last digit of a^b.
Examples: 
 

Input  : 3 10
Output : 9

Input  : 6 2
Output : 6

Input  : 150 53
Output : 0

After taking few examples, we can notice below pattern.
 

Number  |  Last digits that repeat in cycle
  1     |     1
  2     |  4, 8, 6, 2
  3     |  9, 7, 1, 3
  4     |  6, 4
  5     |  5
  6     |  6
  7     |  9, 3, 1, 7
  8     |  4, 2, 6, 8
  9     |  1, 9

In the given table, we can see that maximum length for cycle repetition is 4. 
Example: 2*2 = 4*2 = 8*2 = 16*2 = 32 last digit in 32 is 2 that means after multiplying 4 times digit repeat itself. So the algorithm is very simple .
Source : Brilliants.org
Algorithm :
 

1. Since number are very large we store them as a string.
2. Take last digit in base a.
3. Now calculate b%4. Here b is very large. 
   • If b%4==0 that means b is completely divisible by 4, so our exponent now will be exp = 4 because by multiplying number 4 times, we get the last digit according to cycle table in above diagram.
   • If b%4!=0 that means b is not completely divisible by 4, so our exponent now will be exp=b%4 because by multiplying number exponent times, we get the last digit according to cycle table in above diagram.
   • Now calculate ldigit = pow( last_digit_in_base, exp ).
   • Last digit of a^b will be ldigit%10.

Below is the implementation of above algorithm. 
 

# Python 3 code to find last digit of a ^ b

import math

# Function to find b % a
def Modulo(a, b) :
    # Initialize result
    mod = 0

    # calculating mod of b with a to make
    # b like 0 <= b < a
    for i in range(0, len(b)) :
        mod = (mod * 10 + (int)(b[i])) % a

    return mod # return modulo


# function to find last digit of a ^ b
def LastDigit(a, b) :
    len_a = len(a)
    len_b = len(b)

    # if a and b both are 0
    if (len_a == 1 and len_b == 1 and b[0] == '0' and a[0] == '0') :
        return 1

    # if exponent is 0
    if (len_b == 1 and b[0]=='0') :
        return 1

    # if base is 0
    if (len_a == 1 and a[0] == '0') :
        return 0

    # if exponent is divisible by 4 that means last
    # digit will be pow(a, 4) % 10.
    # if exponent is not divisible by 4 that means last
    # digit will be pow(a, b % 4) % 10
    if((Modulo(4, b) == 0)) :
        exp = 4
    else : 
        exp = Modulo(4, b)

    # Find last digit in 'a' and compute its exponent
    res = math.pow((int)(a[len_a - 1]), exp)

    # Return last digit of result
    return res % 10
    

# Driver program to run test case
a = ['1', '1', '7']
b = ['3']
print(LastDigit(a, b))

# This code is contributed to Nikita Tiwari.

Output : 

3

This article is contributed by Shashank Mishra ( Gullu ). This article is reviewed by team geeksforgeeks. 
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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