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Find Last Digit of a^b for Large Numbers

  • Difficulty Level : Medium
  • Last Updated : 13 Apr, 2021

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. 
 



C++




// C++ code to find last digit of a^b
#include <bits/stdc++.h>
using namespace std;
 
// Function to find b % a
int Modulo(int a, char b[])
{
    // Initialize result
    int mod = 0;
 
    // calculating mod of b with a to make
    // b like 0 <= b < a
    for (int i = 0; i < strlen(b); i++)
        mod = (mod * 10 + b[i] - '0') % a;
 
    return mod; // return modulo
}
 
// function to find last digit of a^b
int LastDigit(char a[], char b[])
{
    int len_a = strlen(a), len_b = strlen(b);
 
    // if a and b both are 0
    if (len_a == 1 && len_b == 1 && b[0] == '0' && a[0] == '0')
        return 1;
 
    // if exponent is 0
    if (len_b == 1 && b[0] == '0')
        return 1;
 
    // if base is 0
    if (len_a == 1 && 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
    int exp = (Modulo(4, b) == 0) ? 4 : Modulo(4, b);
 
    // Find last digit in 'a' and compute its exponent
    int res = pow(a[len_a - 1] - '0', exp);
 
    // Return last digit of result
    return res % 10;
}
 
// Driver program to run test case
int main()
{
    char a[] = "117", b[] = "3";
    cout << LastDigit(a, b);
    return 0;
}

Java




// Java code to find last digit of a^b
import java.io.*;
import java.math.*;
 
class GFG {
 
    // Function to find b % a
    static int Modulo(int a, char b[])
    {
        // Initialize result
        int mod = 0;
 
        // calculating mod of b with a to make
        // b like 0 <= b < a
        for (int i = 0; i < b.length; i++)
            mod = (mod * 10 + b[i] - '0') % a;
 
        return mod; // return modulo
    }
 
    // Function to find last digit of a^b
    static int LastDigit(char a[], char b[])
    {
        int len_a = a.length, len_b = b.length;
 
        // if a and b both are 0
        if (len_a == 1 && len_b == 1 && b[0] == '0' && a[0] == '0')
            return 1;
 
        // if exponent is 0
        if (len_b == 1 && b[0] == '0')
            return 1;
 
        // if base is 0
        if (len_a == 1 && 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
        int exp = (Modulo(4, b) == 0) ? 4 : Modulo(4, b);
 
        // Find last digit in 'a' and compute its exponent
        int res = (int)(Math.pow(a[len_a - 1] - '0', exp));
 
        // Return last digit of result
        return res % 10;
    }
 
    // Driver program to run test case
    public static void main(String args[]) throws IOException
    {
        char a[] = "117".toCharArray(), b[] = { '3' };
        System.out.println(LastDigit(a, b));
    }
}
 
// This code is contributed by Nikita Tiwari.

Python3 


# 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.


C#




// C# code to find last digit of a^b.
using System;
 
class GFG {
 
    // Function to find b % a
    static int Modulo(int a, char[] b)
    {
         
        // Initialize result
        int mod = 0;
 
        // calculating mod of b with a
        // to make b like 0 <= b < a
        for (int i = 0; i < b.Length; i++)
            mod = (mod * 10 + b[i] - '0') % a;
 
        // return modulo
        return mod;
    }
 
    // Function to find last digit of a^b
    static int LastDigit(char[] a, char[] b)
    {
        int len_a = a.Length, len_b = b.Length;
 
        // if a and b both are 0
        if (len_a == 1 && len_b == 1 &&
                   b[0] == '0' && a[0] == '0')
            return 1;
 
        // if exponent is 0
        if (len_b == 1 && b[0] == '0')
            return 1;
 
        // if base is 0
        if (len_a == 1 && 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
        int exp = (Modulo(4, b) == 0) ? 4
                            : Modulo(4, b);
 
        // Find last digit in 'a' and
        // compute its exponent
        int res = (int)(Math.Pow(a[len_a - 1]
                                - '0', exp));
 
        // Return last digit of result
        return res % 10;
    }
 
    // Driver program to run test case
    public static void Main()
    {
         
        char[] a = "117".ToCharArray(),
        b = { '3' };
         
        Console.Write(LastDigit(a, b));
    }
}
 
// This code is contributed by nitin mittal.

PHP




<?php
// php code to find last digit of a^b
 
// Function to find b % a
function Modulo($a, $b)
{
     
    // Initialize result
    $mod = 0;
 
    // calculating mod of b with a to make
    // b like 0 <= b < a
    for ($i = 0; $i < strlen($b); $i++)
        $mod = ($mod * 10 + $b[$i] - '0') % $a;
 
    return $mod; // return modulo
}
 
// function to find last digit of a^b
function LastDigit($a, $b)
{
    $len_a = strlen($a); $len_b = strlen($b);
 
    // if a and b both are 0
    if ($len_a == 1 && $len_b == 1 &&
                $b[0] == '0' && $a[0] == '0')
        return 1;
 
    // if exponent is 0
    if ($len_b == 1 && $b[0] == '0')
        return 1;
 
    // if base is 0
    if ($len_a == 1 && $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
    $exp = (Modulo(4, $b) == 0) ? 4 :
                              Modulo(4, $b);
 
    // Find last digit in 'a' and compute
    // its exponent
    $res = pow($a[$len_a - 1] - '0', $exp);
 
    // Return last digit of result
    return $res % 10;
}
 
// Driver program to run test case
$a = "117";
$b = "3";
echo LastDigit($a, $b);
 
// This code is contributed by nitin mittal.
?>

Javascript




<script>
 
// Javascript code to find last digit of a^b
 
// Function to find b % a
function Modulo(a, b)
{
     
    // Initialize result
    let mod = 0;
 
    // calculating mod of b with a to make
    // b like 0 <= b < a
    for (let i = 0; i < b.length; i++)
        mod = (mod * 10 + b[i] - '0') % a;
 
    return mod; // return modulo
}
 
// function to find last digit of a^b
function LastDigit(a, b)
{
    let len_a = a.length;
    let len_b = b.length;
 
    // if a and b both are 0
    if (len_a == 1 && len_b == 1 &&
                b[0] == '0' && a[0] == '0')
        return 1;
 
    // if exponent is 0
    if (len_b == 1 && b[0] == '0')
        return 1;
 
    // if base is 0
    if (len_a == 1 && 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
    exp = (Modulo(4, b) == 0) ? 4 :
                            Modulo(4, b);
 
    // Find last digit in 'a' and compute
    // its exponent
    res = Math.pow(a[len_a - 1] - '0', exp);
 
    // Return last digit of result
    return res % 10;
}
 
// Driver program to run test case
let a = "117";
let b = "3";
document.write(LastDigit(a, b));
 
// This code is contributed by _saurabh_jaiswal
 
</script>

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.
 

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.




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