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.
// 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;
} |
// 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 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. |
// 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 // 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. ?> |
<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> |
6
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++ 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 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 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# 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 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. ?> |
<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> |
3
Time Complexity: O(log n)
Auxiliary Space: O(1)
Approach: Binomial Expansion method
Here are the steps to find the unit digit of x raised to power y using the Binomial Expansion method:
1. Handle special cases:
If y is 0, return 1 as any number raised to power 0 is 1.
If x is 0, return 0 as any number raised to power 0 is 1 and the unit digit of 0 is 0.
2. Calculate the y-th term in the expansion of (x+10)^y using the binomial theorem:
The y-th term in the expansion is given by: C(y, 0)x^y10^0 + C(y, 1)*x^(y-1)*10^1 + … + C(y, y)x^010^y
Here, C(y, k) represents the binomial coefficient, which is equal to y! / (k! * (y-k)!).
We only need to calculate the last term in this expansion, which is C(y, y)x^010^y.
3. Find the unit digit of the y-th term:
The unit digit of the y-th term is the same as the last digit of the y-th term.
We can find the last digit of the y-th term by taking the remainder of the term when divided by 10.
4. Return the unit digit found in step 3 as the result.
#include <iostream> #include <cmath> using namespace std;
int unit_digit( int x, int y) {
if (y == 0) {
return 1;
}
if (x == 0) {
return 0;
}
int term = pow (x + 10, y);
int last_digit = term % 10;
return last_digit;
} int main() {
cout << unit_digit(2, 1) << endl; // Output: 2
cout << unit_digit(4, 2) << endl; // Output: 6
return 0;
} // This is contributed by uppalasridevi |
import java.lang.Math;
public class Main {
public static int unitDigit( int x, int y) {
if (y == 0 ) {
return 1 ;
}
if (x == 0 ) {
return 0 ;
}
int term = ( int ) Math.pow(x + 10 , y);
int lastDigit = term % 10 ;
return lastDigit;
}
public static void main(String[] args) {
System.out.println(unitDigit( 2 , 1 )); // Output: 2
System.out.println(unitDigit( 4 , 2 )); // Output: 6
}
} |
def unit_digit(x, y):
if y = = 0 :
return 1
if x = = 0 :
return 0
term = (x + 10 ) * * y
last_digit = term % 10
return last_digit
# Using the binomial theorem method print (unit_digit( 2 , 1 )) # Output: 2
# Using the binomial theorem method print (unit_digit( 4 , 2 )) # Output: 6
# This is contributed by uppalasridevi |
// C# Code for the above approach using System;
public class MainClass {
// Function to find the unit digit
public static int UnitDigit( int x, int y)
{
if (y == 0) {
return 1;
}
if (x == 0) {
return 0;
}
int term = ( int )Math.Pow(x + 10, y);
int lastDigit = term % 10;
return lastDigit;
}
// Driver Code
public static void Main( string [] args)
{
Console.WriteLine(UnitDigit(2, 1)); // Output: 2
Console.WriteLine(UnitDigit(4, 2)); // Output: 6
}
} |
function unitDigit(x, y) {
if (y == 0) {
return 1;
}
if (x == 0) {
return 0;
}
let term = Math.pow(x + 10, y);
let lastDigit = term % 10;
return lastDigit;
} console.log(unitDigit(2, 1)); // Output: 2
console.log(unitDigit(4, 2)); // Output: 6
|
2 6
The time complexity is O(log y), where y is the input variable
The auxiliary space also 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