Given a positive integer N, The task is to find the value of the 3rd digit from the last (right-most) of 5N.
Examples:
Input : N = 6 Output : 6 Explanation : Value of 56 = 15625. Input : N = 3 Output : 1 Explanation : Value of 53 = 125.
Approach: Before moving to the actual approach, some facts regarding number theory are listed below as:
- 53 is the smallest 3-digit number which is a power of 5.
- As 125 * 5 = 625, this concludes that multiple of numbers (ending with 125) with 5 always construct 625 as of the last three digits of the result.
- Again as, 625 * 5 = 3125, this concludes that multiple numbers (ending with 625) with 5 always construct 125 as the last three digits of the result.
Hence, the final general solution is :
Case 1: if n < 3, answer = 0.
Case 2: if n >= 3 and is even, answer = 6.
Case 3: if n >= 3 and is odd, answer = 1.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std; // Function to find the element int findThirdDigit( int n) { // if n < 3 if (n < 3) return 0; // If n is even return 6 // If n is odd return 1 return n & 1 ? 1 : 6; } // Driver code int main() { int n = 7; cout << findThirdDigit(n); return 0; } |
Java
// Java implementation of the // above approach import java.io.*; class GFG { // Function to find the element static int findThirdDigit( int n) { // if n < 3 if (n < 3 ) return 0 ; // If n is even return 6 // If n is odd return 1 return (n & 1 ) > 0 ? 1 : 6 ; } // Driver code public static void main(String args[]) { int n = 7 ; System.out.println(findThirdDigit(n)); } } // This code is contributed // by Akanksha Rai |
Python3
# Python3 implementation of the # above approach # Function to find the element def findThirdDigit(n): # if n < 3 if n < 3 : return 0 # If n is even return 6 # If n is odd return 1 return 1 if n and 1 else 6 # Driver code n = 7 print (findThirdDigit(n)) # This code is contributed # by Shrikant13 |
C#
// C# implementation of the above approach using System; class GFG { // Function to find the element static int findThirdDigit( int n) { // if n < 3 if (n < 3) return 0; // If n is even return 6 // If n is odd return 1 return (n & 1)>0 ? 1 : 6; } // Driver code static void Main() { int n = 7; Console.WriteLine(findThirdDigit(n)); } } // This code is contributed by mits |
PHP
<?php // PHP implementation of the above approach // Function to find the element function findThirdDigit( $n ) { // if n < 3 if ( $n < 3) return 0; // If n is even return 6 // If n is odd return 1 return $n & 1 ? 1 : 6; } // Driver code $n = 7; echo findThirdDigit( $n ); // This code contributed by // PrinciRaj1992 ?> |
Javascript
<script> // Javascript implementation of the above approach // Function to find the element function findThirdDigit(n) { // if n < 3 if (n < 3) return 0; // If n is even return 6 // If n is odd return 1 return n & 1 ? 1 : 6; } // Driver code var n = 7; document.write( findThirdDigit(n)); </script> |
1
Time Complexity: O(1)
Auxiliary Space: O(1), since no extra space has been taken.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!