Given a five-digit number N., The task is to find the last five digits of the given number raised to the power of 5 after modifying it by arranging the digits as:
first digit, third digit, fifth digit, fourth digit, second digit.
Examples:
Input : N = 12345
Output : 71232
Explanation :
After modification the number becomes 13542. (13542)5 is
455422043125550171232
Input : N = 10000
Output : 00000
Approach: In this problem, just implementation of the actions described in the statement is required. However, there are two catches in this problem.
The first catch is that the fifth power of a five-digit number cannot be represented by a 64-bit integer. But we do not actually need the fifth power, we need the fifth power modulo 105. And mod operation can be applied after each multiplication.
The second catch is that you need to output five digits, not the fifth power modulo 105. The difference is when the fifth digit from the end is zero. To output, a number with the leading zero one can either use corresponding formatting (%05d in printf) or extract digits and output them one by one.
Below is the implementation of the above approach :
C++
// CPP program to find last five digits// of a five digit number raised to power five#include <bits/stdc++.h>using namespace std;// Function to find the last five digits// of a five digit number raised to power fiveint lastFiveDigits(int n){ n = (n / 10000) * 10000 + ((n / 100) % 10) * 1000 + (n % 10) * 100 + ((n / 10) % 10) * 10 + (n / 1000) % 10; long long ans = 1; for (int i = 0; i < 5; i++) { ans *= n; ans %= 100000; } printf("%05d", ans);}// Driver codeint main(){ int n = 12345; lastFiveDigits(n); return 0;} |
Java
// Java program to find last five digits// of a five digit number raised to power fiveclass GfG { // Function to find the last five digits // of a five digit number raised to power five static void lastFiveDigits(int n) { n = (n / 10000) * 10000 + ((n / 100) % 10) * 1000 + (n % 10) * 100 + ((n / 10) % 10) * 10 + (n / 1000) % 10; int ans = 1; for (int i = 0; i < 5; i++) { ans *= n; ans %= 100000; } System.out.println(ans); } // Driver code public static void main(String[] args) { int n = 12345; lastFiveDigits(n); }} |
Python3
# Python3 program to find last five digits# of a five digit number raised to power five# Function to find the last five digits# of a five digit number raised to power fivedef lastFiveDigits(n): n = ((int)(n / 10000) * 10000 + ((int)(n / 100) % 10) * 1000 + (n % 10) * 100 + ((int)(n / 10) % 10) * 10 + (int)(n / 1000) % 10) ans = 1 for i in range(5): ans *= n ans %= 100000 print(ans)# Driver codeif __name__ == '__main__': n = 12345 lastFiveDigits(n)# This code contributed by PrinciRaj1992 |
C#
// C# program to find last five // digits of a five digit number// raised to power five using System;class GFG{ // Function to find the last // five digits of a five digit // number raised to power five public static void lastFiveDigits(int n) { n = (n / 10000) * 10000 + ((n / 100) % 10) * 1000 + (n % 10) * 100 + ((n / 10) % 10) * 10 + (n / 1000) % 10; int ans = 1; for (int i = 0; i < 5; i++) { ans *= n; ans %= 100000; } Console.WriteLine(ans); } // Driver code public static void Main(string[] args) { int n = 12345; lastFiveDigits(n); }}// This code is contributed// by Shrikant13 |
Javascript
<script>// JavaScript program to find last five digits// of a five digit number raised to power five// Function to find the last five digits// of a five digit number raised to power fivefunction lastFiveDigits(n){ n = (Math.floor(n / 10000)) * 10000 + (Math.floor(n / 100) % 10) * 1000 + (n % 10) * 100 + (Math.floor(n / 10) % 10) * 10 + Math.floor(n / 1000) % 10; let ans = 1; for (let i = 0; i < 5; i++) { ans *= n; ans %= 100000; } document.write(ans);}// Driver code let n = 12345; lastFiveDigits(n);// This code is contributed by Manoj.</script> |
PHP
<?php// PHP program to find last five digits// of a five digit number raised to power five// Function to find the last five digits// of a five digit number raised to power fivefunction lastFiveDigits($n){ $n = (int)($n / 10000) * 10000 + ((int)($n / 100) % 10) * 1000 + ($n % 10) * 100 + ((int)($n / 10) % 10) * 10 + (int)($n / 1000) % 10; $ans = 1; for ($i = 0; $i < 5; $i++) { $ans *= $n; $ans %= 100000; } echo $ans;}// Driver code$n = 12345;lastFiveDigits($n);// This code is contributed// by Akanksha Rai?> |
Output
71232
Time Complexity: O(1)
Auxiliary Space: O(1)
Approach: Modified Digit Arrangement and Modulo Operation”
The “Modified Digit Arrangement and Modulo Operation” approach to finding the last five digits of a five-digit number raised to power five consists of the following steps:
- Modify the number as per the given arrangement of digits.
- Calculate the power of 5 of the modified number.
- Take modulo with 100000 (10^5) to get the last five digits.
The key idea behind this approach is to use the given arrangement of digits to modify the original number in such a way that the resulting number has the same last five digits as the original number when raised to power five. Then, by taking modulo with 100000, we can obtain the last five digits of the resulting number.
C++
#include <iostream>// Function to calculate (base^exponent) % modulus using modular exponentiationint power(int base, int exponent, int modulus) { if (exponent == 0) return 1; long long result = 1; long long x = base % modulus; while (exponent > 0) { if (exponent % 2 == 1) result = (result * x) % modulus; x = (x * x) % modulus; exponent /= 2; } return result;}// Function to find the last five digits of the number N after performing the given operationsint findLastFiveDigits(int N) { // Extract individual digits of the number int digit1 = N / 10000; int digit2 = (N / 1000) % 10; int digit3 = (N / 100) % 10; int digit4 = (N / 10) % 10; int digit5 = N % 10; // Modify the number as per the given arrangement of digits int new_N = digit1 * 10000 + digit3 * 1000 + digit5 * 100 + digit4 * 10 + digit2; std::cout << "Modified number: " << new_N << std::endl; // Calculate the power of 5 of the modified number using modular exponentiation int powerVal = power(new_N, 5, 100000); std::cout << "Power of 5: " << powerVal << std::endl; // Take modulo with 100000 to get the last five digits int last_five_digits = powerVal % 100000; std::cout << "Last five digits: " << last_five_digits << std::endl; return last_five_digits;}int main() { std::cout << findLastFiveDigits(12345) << std::endl; std::cout << findLastFiveDigits(10000) << std::endl; // This code is contributed by Shivam Tiwari return 0;} |
Java
public class GFG { public static void main(String[] args) { System.out.println(findLastFiveDigits(12345)); System.out.println(findLastFiveDigits(10000)); } public static int findLastFiveDigits(int N) { // Modify the number as per the given arrangement of // digits int new_N = Integer.parseInt( "" + Integer.toString(N).charAt(0) + Integer.toString(N).charAt(2) + Integer.toString(N).charAt(4) + Integer.toString(N).charAt(3) + Integer.toString(N).charAt(1)); System.out.println("Modified number: " + new_N); // Calculate the power of 5 of the modified number long power = (long)Math.pow(new_N, 5); System.out.println("Power of 5: " + power); // Take modulo with 100000 to get the last five // digits int last_five_digits = (int)(power % 100000); System.out.println("Last five digits: " + last_five_digits); return last_five_digits; }} |
Python3
def find_last_five_digits(N): # Modify the number as per the given arrangement of digits new_N = int(str(N)[0] + str(N)[2] + str(N)[4] + str(N)[3] + str(N)[1]) print(f"Modified number: {new_N}") # Calculate the power of 5 of the modified number power = new_N ** 5 print(f"Power of 5: {power}") # Take modulo with 100000 to get the last five digits last_five_digits = power % 100000 print(f"Last five digits: {last_five_digits}") return last_five_digitsprint(find_last_five_digits(12345))print(find_last_five_digits(10000)) |
C#
using System;public class GFG { public static void Main(string[] args) { Console.WriteLine(findLastFiveDigits(12345)); Console.WriteLine(findLastFiveDigits(10000)); } public static int findLastFiveDigits(int N) { // Modify the number as per the given arrangement of // digits int new_N = int.Parse( "" + N.ToString()[0] + N.ToString()[2] + N.ToString()[4] + N.ToString()[3] + N.ToString()[1]); Console.WriteLine("Modified number: " + new_N); // Calculate the power of 5 of the modified number long power = (long)Math.Pow(new_N, 5); Console.WriteLine("Power of 5: " + Math.Abs(power)); // Take modulo with 100000 to get the last five // digits int last_five_digits = (int)(power % 100000); Console.WriteLine("Last five digits: " + Math.Abs(last_five_digits)); return Math.Abs(last_five_digits); }}// This code is contributed by shiv1o43g |
Javascript
function findLastFiveDigits(N) { // Modify the number as per the given arrangement of digits let new_N = parseInt(N.toString()[0] + N.toString()[2] + N.toString()[4] + N.toString()[3] + N.toString()[1]); console.log("Modified number: " + new_N); // Calculate the power of 5 of the modified number let power = Math.pow(new_N, 5); console.log("Power of 5: " + power); // Take modulo with 100000 to get the last five digits let lastFiveDigits = power % 100000; console.log("Last five digits: " + lastFiveDigits); return lastFiveDigits;}console.log(findLastFiveDigits(12345));console.log(findLastFiveDigits(10000)); |
Output
Modified number: 13542 Power of 5: 455422043125550171232 Last five digits: 71232 71232 Modified number: 10000 Power of 5: 100000000000000000000 Last five digits: 0 0
The time complexity is O(1)
The auxiliary space is also O(1)
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