Find unit digit of x raised to power y
Given two numbers x and y, find unit digit of xy.
Examples :
Input : x = 2, y = 1 Output : 2 Explanation 2^1 = 2 so units digit is 2. Input : x = 4, y = 2 Output : 6 Explanation 4^2 = 16 so units digit is 6.
Method 1 (Simple) Compute value of xy and find its last digit. This method causes overflow for slightly larger values of x and y.
Method 2 (Efficient)
1) Find last digit of x.
2) Compute x^y under modulo 10 and return its value.
C++
// Efficient C++ program to // find unit digit of x^y. #include <bits/stdc++.h> using namespace std; // Returns unit digit of x // raised to power y int unitDigitXRaisedY( int x, int y) { // Initialize result as 1 to // handle case when y is 0. int res = 1; // One by one multiply with x // mod 10 to avoid overflow. for ( int i = 0; i < y; i++) res = (res * x) % 10; return res; } // Driver program int main() { cout << unitDigitXRaisedY(4, 2); return 0; } |
Java
// Efficient Java program to find // unit digit of x^y. import java.io.*; class GFG { // Returns unit digit of x raised to power y static int unitDigitXRaisedY( int x, int y) { // Initialize result as 1 to // handle case when y is 0. int res = 1 ; // One by one multiply with x // mod 10 to avoid overflow. for ( int i = 0 ; i < y; i++) res = (res * x) % 10 ; return res; } // Driver program public static void main(String args[]) throws IOException { System.out.println(unitDigitXRaisedY( 4 , 2 )); } } // This code is contributed by Nikita Tiwari. |
Python3
# Python3 code to find # unit digit of x^y. # Returns unit digit of # x raised to power y def unitDigitXRaisedY( x , y ): # Initialize result as 1 to # handle case when y is 0. res = 1 # One by one multiply with x # mod 10 to avoid overflow. for i in range (y): res = (res * x) % 10 return res # Driver program print ( unitDigitXRaisedY( 4 , 2 )) # This code is contributed by Abhishek Sharma44. |
C#
// Efficient Java program to find // unit digit of x^y. using System; class GFG { // Returns unit digit of x raised to power y static int unitDigitXRaisedY( int x, int y) { // Initialize result as 1 to // handle case when y is 0. int res = 1; // One by one multiply with x // mod 10 to avoid overflow. for ( int i = 0; i < y; i++) res = (res * x) % 10; return res; } // Driver program public static void Main() { Console.WriteLine(unitDigitXRaisedY(4, 2)); } } // This code is contributed by vt_m. |
PHP
<?php // Efficient PHP program to // find unit digit of x^y. // Returns unit digit of x // raised to power y function unitDigitXRaisedY( $x , $y ) { // Initialize result as 1 to // handle case when y is 0. $res = 1; // One by one multiply with x // mod 10 to avoid overflow. for ( $i = 0; $i < $y ; $i ++) $res = ( $res * $x ) % 10; return $res ; } // Driver Code echo (unitDigitXRaisedY(4, 2)); // This code is contributed by Ajit. ?> |
Javascript
<script> // Efficient Javascript program to // find unit digit of x^y. // Returns unit digit of x // raised to power y function unitDigitXRaisedY(x, y) { // Initialize result as 1 to // handle case when y is 0. let res = 1; // One by one multiply with x // mod 10 to avoid overflow. for (let i = 0; i < y; i++) res = (res * x) % 10; return res; } // Driver Code document.write(unitDigitXRaisedY(4, 2)); // This code is contributed by _saurabh_jaiswal. </script> |
Output :
6
Time Complexity: O(y), where y is the power
Auxiliary Space: O(1), as no extra space is required
Further Optimizations: We can compute modular power in Log y.
Method 3 (Direct based on cyclic nature of last digit)
This method depends on the cyclicity with the last digit of x that is
x | power 2 | power 3 | power 4 | Cyclicity 0 | .................................. | .... repeat with 0 1 | .................................. | .... repeat with 1 2 | 4 | 8 | 6 | .... repeat with 2 3 | 9 | 7 | 1 | .... repeat with 3 4 | 6 |....................... | .... repeat with 4 5 | .................................. | .... repeat with 5 6 | .................................. | .... repeat with 6 7 | 9 | 3 | 1 | .... repeat with 7 8 | 4 | 2 | 6 | .... repeat with 8 9 | 1 | ...................... | .... repeat with 9
So here we directly mod the power y with 4 because this is the last power after this all number’s repetition start
after this we simply power with number x last digit then we get the unit digit of produced number.
C++
// C++ code to find the unit digit of x // raised to power y. #include<iostream> #include<math.h> using namespace std; // find unit digit int unitnumber( int x, int y) { // Get last digit of x x = x % 10; // Last cyclic modular value if (y!=0) y = y % 4 + 4; // here we simply return the // unit digit or the power // of a number return ((( int )( pow (x, y))) % 10); } int main() { int x = 133, y = 5; // get unit digit number here we pass // the unit digit of x and the last // cyclicity number that is y%4 cout << unitnumber(x, y); return 0; } |
Java
// Java code to find the unit // digit of x raised to power y. import java.io.*; import java.util.*; class GFG { // find unit digit static int unitnumber( int x, int y) { // Get last digit of x x = x % 10 ; // Last cyclic modular value if (y!= 0 ) y = y % 4 + 4 ; // here we simply return the // unit digit or the power // of a number return ((( int )(Math.pow(x, y))) % 10 ); } public static void main (String[] args) { int x = 133 , y = 5 ; // get unit digit number here we pass // the unit digit of x and the last // cyclicity number that is y%4 System.out.println(unitnumber(x, y)); } } // This code is contributed by Gitanjali. |
Python3
# Python3 code to find the unit # digit of x raised to power y. import math # Find unit digit def unitnumber(x, y): # Get last digit of x x = x % 10 # Last cyclic modular value if y! = 0 : y = y % 4 + 4 # Here we simply return # the unit digit or the # power of a number return ((( int )(math. pow (x, y))) % 10 ) # Driver code x = 133 ; y = 5 # Get unit digit number here we pass # the unit digit of x and the last # cyclicity number that is y%4 print (unitnumber(x, y)) # This code is contributed by Gitanjali. |
C#
// C# code to find the unit // digit of x raised to power y. using System; class GFG { // find unit digit static int unitnumber( int x, int y) { // Get last digit of x x = x % 10; // Last cyclic modular value if (y!=0) y = y % 4 + 4; // here we simply return the // unit digit or the power // of a number return ((( int )(Math.Pow(x, y))) % 10); } // Driver code public static void Main () { int x = 133, y = 5; // get unit digit number here we pass // the unit digit of x and the last // cyclicity number that is y%4 Console.WriteLine(unitnumber(x, y)); } } // This code is contributed by vt_m. |
PHP
<?php // PHP code to find the unit // digit of x raised to power y. // find unit digit function unitnumber( $x , $y ) { // Get last digit of x $x = $x % 10; // Last cyclic modular value if ( $y !=0) $y = $y % 4 + 4; // here we simply return the // unit digit or the power // of a number return (((int)(pow( $x , $y ))) % 10); } // Driver code $x = 133; $y = 5; // get unit digit number here we pass // the unit digit of x and the last // cyclicity number that is y%4 echo (unitnumber( $x , $y )); // This code is contributed by Ajit. ?> |
Javascript
<script> // Javascript code to find the unit // digit of x raised to power y. // find unit digit function unitnumber(x, y) { // Get last digit of x x = x % 10; // Last cyclic modular value if (y != 0) y = y % 4 + 4; // here we simply return the // unit digit or the power // of a number return ((parseInt(Math.pow(x, y))) % 10); } // Driver code let x = 133; let y = 5; // get unit digit number here we pass // the unit digit of x and the last // cyclicity number that is y%4 document.write(unitnumber(x, y)); // This code is contributed by _saurabh_jaiswal. </script> |
Output :
3
Time Complexity: O(log n)
Auxiliary Space: O(1)
Thanks to DevanshuAgarwal for suggesting above solution.
How to handle large numbers?
Efficient method for Last Digit Of a^b for Large Numbers