Last Digit of a^b

Given two integers a and b in the form of strings. Return the last digit of a^b.

Examples:

Input: a = "3", b = "10"
Output: 9
Explanation: 3¹⁰ = 59049. The last digit is 9.

Input: a = "6", b = "2"
Output: 6
Explanation: 6² = 36. The last digit is 6.

Using Cyclicity of Last Digit (Mod 10) - O(m) Time O(1) Space

The last digit of powers follows a repeating cycle (maximum length = 4). Instead of computing large powers, we reduce the exponent using modulo 4 and use only the last digit of the base.

Here are few examples numbers and last digits of their powers

• 1: 1, 1, 1, 1, 1, 1, .... [Cycle Length is 1]
• 2: 4, 8, 6, 2, 4, 8, 6, 2,... [Cycle Length is 4]
• 3: 9, 7, 1, 3, 9, 7, 1, 3,... [Cycle Length is 4]
• 4: 6, 4, 6, 4, 6, 4, 6, 4, ...... [Cycle Length is 2]
• 5: 5, 5, 5, 5, 5, ... [Cycle Length is 1]
• 6: 6, 6, 6, 6, 6...... [Cycle Length is 1]
• 7: 9, 3, 1, 7, 9, 3, 1, 7,...... [Cycle Length is 4]
• 8: 4, 2, 6, 8, 4, 2, 6, 8, ..... [Cycle Length is 4]
• 9: 1, 9, 1, 9, 1, 9...... [Cycle Length is 2]

Algorithm:

• Extract the last digit of base a.
• Compute b % 4 using string-based modulo (since b is large).
• If b % 4 == 0, take the exponent as 4, otherwise b % 4.
• Compute power using reduced exponent.
• Return the last digit of the result (% 10).

Why does this work?

• Any number’s last digit depends o (a^b) mod 10. The possible last digits are 0–9. When you keep multiplying, results must eventually repeat (finite possibilities)
• If the base 'a' is coprime with 10 (i.e., last digit is 1, 3, 7, or 9), then (a^4) mod 10 is 1. After every 4 powers, the pattern resets. so last digits repeat every 4.
• If the base is not coprime with 10 (i.e., last digit is 2, 4, 5, 6 or 8), then we can find the pattern case by case basis as there are finite possibilities. For example for 2, 5 or 6, it is always the same digit.

#include <bits/stdc++.h>
using namespace std;

int modulo(int d, string &b)
{
    int mod = 0;

    // calculate (b mod d) to make
    // b in range 0 <= b < d
    for (int i = 0; i < b.length(); i++)
        mod = (mod * 10 + b[i] - '0') % d;

    return mod;
}

int getLastDigit(string &a, string &b)
{

    int n = a.length(), m = b.length();

    // exponent is 0
    if (m == 1 && b[0] == '0')
        return 1;

    // base is 0
    if (n == 1 && a[0] == '0')
        return 0;

    // if exponent is divisible by 4 that means last
    // digit will be pow(a, 4) % 10
    // otherwise last digit will be pow(a, b%4) % 10
    int exp = (modulo(4, b) == 0) ? 4 : modulo(4, b);
    int res = pow(a[n - 1] - '0', exp);
    return res % 10;
}

int main()
{
    string a = "3", b = "10";
    cout << getLastDigit(a, b);
    return 0;
}

import java.util.*;

public class GFG {
    int modulo(int d, String b)
    {
        int mod = 0;

        // calculate (b mod d) to make
        // b in range 0 <= b < d
        for (int i = 0; i < b.length(); i++)
            mod = (mod * 10 + b.charAt(i) - '0') % d;

        return mod;
    }

    int getLastDigit(String a, String b)
    {

        int n = a.length(), m = b.length();

        // exponent is 0
        if (m == 1 && b.charAt(0) == '0')
            return 1;

        // base is 0
        if (n == 1 && a.charAt(0) == '0')
            return 0;

        // if exponent is divisible by 4 that means last
        // digit will be pow(a, 4) % 10
        // otherwise last digit will be pow(a, b%4) % 10
        int exp = (modulo(4, b) == 0) ? 4 : modulo(4, b);
        int res = (int)Math.pow(a.charAt(n - 1) - '0', exp);
        return res % 10;
    }

    public static void main(String[] args)
    {
        GfG m = new GfG();
        String a = "3", b = "10";
        System.out.println(m.getLastDigit(a, b));
    }
}

def modulo(d, b):
    mod = 0

    # calculate (b mod d) to make
    # b in range 0 <= b < d
    for i in range(len(b)):
        mod = (mod * 10 + int(b[i])) % d

    return mod


def getLastDigit(a, b):

    n = len(a)
    m = len(b)

    # exponent is 0
    if m == 1 and b[0] == '0':
        return 1

    # base is 0
    if n == 1 and a[0] == '0':
        return 0

    # if exponent is divisible by 4 that means last
    # digit will be pow(a, 4) % 10
    # otherwise last digit will be pow(a, b%4) % 10
    exp = 4 if modulo(4, b) == 0 else modulo(4, b)
    res = pow(int(a[n - 1]), exp)
    return res % 10


if __name__ == "__main__":
    a = "3"
    b = "10"
    print(getLastDigit(a, b))

public class GFG {
    public int modulo(int d, string b)
    {
        int mod = 0;

        // calculate (b mod d) to make
        // b in range 0 <= b < d
        for (int i = 0; i < b.Length; i++)
            mod = (mod * 10 + b[i] - '0') % d;

        return mod;
    }

    public int getLastDigit(string a, string b)
    {
        int n = a.Length, m = b.Length;

        // exponent is 0
        if (m == 1 && b[0] == '0')
            return 1;

        // base is 0
        if (n == 1 && a[0] == '0')
            return 0;

        int mod = modulo(4, b);
        int exp = (mod == 0) ? 4 : mod;

        int res = (int)Math.Pow(a[n - 1] - '0', exp);
        return res % 10;
    }

    public static void Main()
    {
        string a = "3", b = "10";
        Console.WriteLine(getLastDigit(a, b));
    }
}

function modulo(d, b)
{
    let mod = 0;

    // calculate (b mod d) to make
    // b in range 0 <= b < d
    for (let i = 0; i < b.length; i++) {
        mod = (mod * 10 + parseInt(b[i])) % d;
    }

    return mod;
}

function getLastDigit(a, b)
{

    let n = a.length, m = b.length;

    // exponent is 0
    if (m == 1 && b[0] == "0")
        return 1;

    // base is 0
    if (n == 1 && a[0] == "0")
        return 0;

    // if exponent is divisible by 4 that means last
    // digit will be pow(a, 4) % 10
    // otherwise last digit will be pow(a, b%4) % 10
    let exp = (modulo(4, b) == 0) ? 4 : modulo(4, b);
    let res = Math.pow(parseInt(a[n - 1]), exp);
    return Math.floor(res % 10);
}

// Driver code
let a = "3", b = "10";
console.log(getLastDigit(a, b));

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