Given two number X and N, the task is to find the last digit of X raised to last digit of N factorial, i.e.
Examples:
Input: X = 5, N = 2
Output: 5
Explanation:
Since, 2! mod 10 = 2
therefore 52 = 25 and the last digit of 25 is 5.
Input: X = 10, N = 4
Output: 0
Explanation:
Since, 4! mod 10 = 24 mod 10 = 4
therefore 104 = 10000 and the last digit of 10000 is 0.
Approach: The most efficient way to solve this problem is to find any pattern in the required last digit, with the help of last digit of N! and last digit of X raised to Y
Below is the various observation of the above-given equation:
- If N = 0 or N = 1, then the last digit is 1 or
respectively. - Since 5! is 120, therefore for N ? 5 the value of (N! mod 10) will be zero.
- Now we are left with digit 2, 3, 4. For this we have:
for N = 2,
N! mod 10 = 2! mod 10 = 2
for N = 3,
N! mod 10 = 3! mod 10 = 6
for N = 4,
N! mod 10 = 4! mod 10 = 24 mod 10 = 4
Now for X2, X4, and X6
we will check that after which nth power of Xn the value of last digit repeats,
i.e, after which nth power of last digit of Xn the value of last digit repeats.
- Below is the table for what power of the last digit from 0 to 9 in any number repeats:
Number | Cyclicity |
---|---|
0 | 1 |
1 | 1 |
2 | 4 |
3 | 4 |
4 | 2 |
5 | 1 |
6 | 1 |
7 | 4 |
8 | 4 |
9 | 2 |
Below are the steps based on the above observations:
- If X is not a multiple of 10 then divide the evaluated value of
by cyclicity of the last digit of X. If remainder(say r) is 0 then do the following: - If the last digit of X is any of 2, 4, 6, or 8 then the answer will be 6.
- If the last digit of X is any of 1, 3, 7, or 9 then the answer will be 1.
- If the last digit of X is 5 then answer will be 5.
- Else if remainder(say r) is a non-zero then answer is
, where ‘l’ is the last digit of X. - Else if X is a multiple of 10 then the answer will be 0 always.
Below is the implementation of the above approach:
// C++ program for the above approach #include<bits/stdc++.h> using namespace std;
// Function to find a^b using // binary exponentiation long power( long a, long b, long c)
{ // Initialise result
long result = 1;
while (b > 0)
{
// If b is odd then,
// multiply result by a
if ((b & 1) == 1)
{
result = (result * a) % c;
}
// b must be even now
// Change b to b/2
b /= 2;
// Change a = a^2
a = (a * a) % c;
}
return result;
} // Function to find the last digit // of the given equation long calculate( long X, long N)
{ int a[10];
// To store cyclicity
int cyclicity[11];
// Store cyclicity from 1 - 10
cyclicity[1] = 1;
cyclicity[2] = 4;
cyclicity[3] = 4;
cyclicity[4] = 2;
cyclicity[5] = 1;
cyclicity[6] = 1;
cyclicity[7] = 4;
cyclicity[8] = 4;
cyclicity[9] = 2;
cyclicity[10] = 1;
// Observation 1
if (N == 0 || N == 1)
{
return (X % 10);
}
// Observation 3
else if (N == 2 || N == 3 || N == 4)
{
long temp = ( long )1e18;
// To store the last digits
// of factorial 2, 3, and 4
a[2] = 2;
a[3] = 6;
a[4] = 4;
// Find the last digit of X
long v = X % 10;
// Step 1
if (v != 0)
{
int u = cyclicity[( int )v];
// Divide a[N] by cyclicity
// of v
int r = a[( int )N] % u;
// If remainder is 0
if (r == 0)
{
// Step 1.1
if (v == 2 || v == 4 ||
v == 6 || v == 8)
{
return 6;
}
// Step 1.2
else if (v == 5)
{
return 5;
}
// Step 1.3
else if (v == 1 || v == 3 ||
v == 7 || v == 9)
{
return 1;
}
}
// If r is non-zero,
// then return (l^r) % 10
else
{
return (power(v, r, temp) % 10);
}
}
// Else return 0
else
{
return 0;
}
}
// Else return 1
return 1;
} // Driver Code int main()
{ // Given Numbers
int X = 18;
int N = 4;
// Function Call
long result = calculate(X, N);
// Print the result
cout << result;
} // This code is contributed by spp____ |
import java.util.*;
class TestClass {
// Function to find a^b using
// binary exponentiation
public static long power( long a, long b, long c) {
// Initialise result
long result = 1 ;
while (b > 0 ) {
// If b is odd then,
// multiply result by a
if ((b & 1 ) == 1 ) {
result = (result * a) % c;
}
// b must be even now
// Change b to b/2
b /= 2 ;
// Change a = a^2
a = (a * a) % c;
}
return result;
}
// Function to find the last digit
// of the given equation
public static long calculate( long X, long N) {
int a[] = new int [ 10 ];
// To store cyclicity
int cyclicity[] = new int [ 11 ];
// Store cyclicity from 1 - 10
cyclicity[ 1 ] = 1 ;
cyclicity[ 2 ] = 4 ;
cyclicity[ 3 ] = 4 ;
cyclicity[ 4 ] = 2 ;
cyclicity[ 5 ] = 1 ;
cyclicity[ 6 ] = 1 ;
cyclicity[ 7 ] = 4 ;
cyclicity[ 8 ] = 4 ;
cyclicity[ 9 ] = 2 ;
cyclicity[ 10 ] = 1 ;
// Observation 1
if (N == 0 || N == 1 ) {
return (X % 10 );
}
// Observation 3
else if (N == 2 || N == 3 || N == 4 ) {
long temp = ( long ) 1e18;
// To store the last digits
// of factorial 2, 3, and 4
a[ 2 ] = 2 ;
a[ 3 ] = 6 ;
a[ 4 ] = 4 ;
// Find the last digit of X
long v = X % 10 ;
// Step 1
if (v != 0 ) {
int u = cyclicity[( int ) v];
// Divide a[N] by cyclicity
// of v
int r = a[( int ) N] % u;
// If remainder is 0
if (r == 0 ) {
// Step 1.1
if (v == 2 || v == 4 || v == 6 || v == 8 ) {
return 6 ;
}
// Step 1.2
else if (v == 5 ) {
return 5 ;
}
// Step 1.3
else if (v == 1 || v == 3 || v == 7 || v == 9 ) {
return 1 ;
}
}
// If r is non-zero,
// then return (l^r) % 10
else {
return (power(v, r, temp) % 10 );
}
}
// Else return 0
else {
return 0 ;
}
}
// Else return 1
return 1 ;
}
// Driver's Code
public static void main(String args[])
throws Exception
{
// Given Numbers
int X = 18 ;
int N = 4 ;
// Function Call
long result = calculate(X, N);
// Print the result
System.out.println(result);
}
} |
# Python3 program for the above approach # Function to find a^b using # binary exponentiation def power(a, b, c):
# Initialise result
result = 1
while (b > 0 ):
# If b is odd then,
# multiply result by a
if ((b & 1 ) = = 1 ):
result = (result * a) % c
# b must be even now
# Change b to b/2
b / / = 2
# Change a = a^2
a = (a * a) % c
return result
# Function to find the last digit # of the given equation def calculate(X, N):
a = 10 * [ 0 ]
# To store cyclicity
cyclicity = 11 * [ 0 ]
# Store cyclicity from 1 - 10
cyclicity[ 1 ] = 1
cyclicity[ 2 ] = 4
cyclicity[ 3 ] = 4
cyclicity[ 4 ] = 2
cyclicity[ 5 ] = 1
cyclicity[ 6 ] = 1
cyclicity[ 7 ] = 4
cyclicity[ 8 ] = 4
cyclicity[ 9 ] = 2
cyclicity[ 10 ] = 1
# Observation 1
if (N = = 0 or N = = 1 ):
return (X % 10 )
# Observation 3
elif (N = = 2 or N = = 3 or N = = 4 ):
temp = 1e18 ;
# To store the last digits
# of factorial 2, 3, and 4
a[ 2 ] = 2
a[ 3 ] = 6
a[ 4 ] = 4
# Find the last digit of X
v = X % 10
# Step 1
if (v ! = 0 ):
u = cyclicity[v]
# Divide a[N] by cyclicity
# of v
r = a[N] % u
# If remainder is 0
if (r = = 0 ):
# Step 1.1
if (v = = 2 or v = = 4 or
v = = 6 or v = = 8 ):
return 6
# Step 1.2
elif (v = = 5 ):
return 5
# Step 1.3
elif (v = = 1 or v = = 3 or
v = = 7 or v = = 9 ):
return 1
# If r is non-zero,
# then return (l^r) % 10
else :
return (power(v, r, temp) % 10 )
# Else return 0
else :
return 0
# Else return 1
return 1
# Driver Code if __name__ = = "__main__" :
# Given numbers
X = 18
N = 4
# Function call
result = calculate(X, N)
# Print the result
print (result)
# This code is contributed by chitranayal |
// C# program for the above approach using System;
using System.Collections.Generic;
class GFG{
// Function to find a^b using // binary exponentiation static long power( long a, long b, long c)
{ // Initialise result
long result = 1;
while (b > 0)
{
// If b is odd then,
// multiply result by a
if ((b & 1) == 1)
{
result = (result * a) % c;
}
// b must be even now
// Change b to b/2
b /= 2;
// Change a = a^2
a = (a * a) % c;
}
return result;
} // Function to find the last digit // of the given equation public static long calculate( long X,
long N)
{ int [] a = new int [10];
// To store cyclicity
int [] cyclicity = new int [11];
// Store cyclicity from 1 - 10
cyclicity[1] = 1;
cyclicity[2] = 4;
cyclicity[3] = 4;
cyclicity[4] = 2;
cyclicity[5] = 1;
cyclicity[6] = 1;
cyclicity[7] = 4;
cyclicity[8] = 4;
cyclicity[9] = 2;
cyclicity[10] = 1;
// Observation 1
if (N == 0 || N == 1)
{
return (X % 10);
}
// Observation 3
else if (N == 2 || N == 3 || N == 4)
{
long temp = ( long )1e18;
// To store the last digits
// of factorial 2, 3, and 4
a[2] = 2;
a[3] = 6;
a[4] = 4;
// Find the last digit of X
long v = X % 10;
// Step 1
if (v != 0)
{
int u = cyclicity[( int )v];
// Divide a[N] by cyclicity
// of v
int r = a[( int )N] % u;
// If remainder is 0
if (r == 0)
{
// Step 1.1
if (v == 2 || v == 4 ||
v == 6 || v == 8)
{
return 6;
}
// Step 1.2
else if (v == 5)
{
return 5;
}
// Step 1.3
else if ( v == 1 || v == 3 ||
v == 7 || v == 9)
{
return 1;
}
}
// If r is non-zero,
// then return (l^r) % 10
else
{
return (power(v, r, temp) % 10);
}
}
// Else return 0
else
{
return 0;
}
}
// Else return 1
return 1;
} // Driver code static void Main()
{ // Given numbers
int X = 18;
int N = 4;
// Function call
long result = calculate(X, N);
// Print the result
Console.Write(result);
} } // This code is contributed by divyeshrabadiya07 |
<script> // JavaScript program for the above approach
// Function to find a^b using
// binary exponentiation
function power(a, b, c) {
// Initialise result
var result = 1;
while (b > 0) {
// If b is odd then,
// multiply result by a
if ((b & 1) == 1) {
result = (result * a) % c;
}
// b must be even now
// Change b to b/2
b /= 2;
// Change a = a^2
a = (a * a) % c;
}
return result;
}
// Function to find the last digit
// of the given equation
function calculate(X, N) {
var a = [...Array(10)];
// To store cyclicity
var cyclicity = [...Array(11)];
// Store cyclicity from 1 - 10
cyclicity[1] = 1;
cyclicity[2] = 4;
cyclicity[3] = 4;
cyclicity[4] = 2;
cyclicity[5] = 1;
cyclicity[6] = 1;
cyclicity[7] = 4;
cyclicity[8] = 4;
cyclicity[9] = 2;
cyclicity[10] = 1;
// Observation 1
if (N == 0 || N == 1) {
return X % 10;
}
// Observation 3
else if (N == 2 || N == 3 || N == 4) {
var temp = 1e18;
// To store the last digits
// of factorial 2, 3, and 4
a[2] = 2;
a[3] = 6;
a[4] = 4;
// Find the last digit of X
var v = X % 10;
// Step 1
if (v != 0) {
var u = cyclicity[parseInt(v)];
// Divide a[N] by cyclicity
// of v
var r = a[parseInt(N)] % u;
// If remainder is 0
if (r == 0)
{
// Step 1.1
if (v == 2 || v == 4 || v == 6 || v == 8) {
return 6;
}
// Step 1.2
else if (v == 5) {
return 5;
}
// Step 1.3
else if (v == 1 || v == 3 || v == 7 || v == 9) {
return 1;
}
}
// If r is non-zero,
// then return (l^r) % 10
else {
return power(v, r, temp) % 10;
}
}
// Else return 0
else {
return 0;
}
}
// Else return 1
return 1;
}
// Driver Code
// Given Numbers
var X = 18;
var N = 4;
// Function Call
var result = calculate(X, N);
// Print the result
document.write(result);
// This code is contributed by rdtank.
</script>
|
Output:
6
Time Complexity: O(logn) because it is using a while loop
Auxiliary Space: O(11)
Approach#2: using for loop
We can first calculate the last digit of N factorial and then take the last digit of X raised to this last digit. We can use modular arithmetic to calculate the last digit of N factorial and to take the last digit of X raised to this value.
Algorithm
1. Calculate the last digit of N factorial using modular arithmetic.
2. Calculate the last digit of X raised to this last digit using modular arithmetic.
3. Return the result.
#include <cmath> #include <iostream> using namespace std;
// This function calculates the last digit of X^N factorial int lastDigit( int X, int N)
{ // Initialize the last digit of N factorial as 1
int lastDigitNFactorial = 1;
// Calculate the last digit of N factorial
for ( int i = 2; i <= N; i++) {
lastDigitNFactorial
= (lastDigitNFactorial * (i % 10)) % 10;
}
// Calculate the last digit of X^(N factorial)
int lastDigitXToLastDigitNFactorial
= ( int ) pow (X % 10, lastDigitNFactorial) % 10;
// Return the result
return lastDigitXToLastDigitNFactorial;
} int main()
{ // Initialize X and N
int X = 18;
int N = 4;
// Print the result of lastDigit(X, N)
cout << lastDigit(X, N) << endl;
return 0;
} |
public class Main {
public static int lastDigit( int X, int N)
{
int lastDigitNFactorial = 1 ;
for ( int i = 2 ; i <= N; i++) {
lastDigitNFactorial
= (lastDigitNFactorial * (i % 10 )) % 10 ;
}
int lastDigitXToLastDigitNFactorial
= ( int )Math.pow(X % 10 , lastDigitNFactorial)
% 10 ;
return lastDigitXToLastDigitNFactorial;
}
public static void main(String[] args)
{
int X = 18 ;
int N = 4 ;
System.out.println(lastDigit(X, N));
}
} |
def last_digit(X, N):
last_digit_n_factorial = 1
for i in range ( 2 , N + 1 ):
last_digit_n_factorial = (last_digit_n_factorial * (i % 10 )) % 10
last_digit_X_to_last_digit_n_factorial = pow (
X % 10 , last_digit_n_factorial, 10 )
return last_digit_X_to_last_digit_n_factorial
X = 18
N = 4
print (last_digit(X, N))
|
using System;
class Program {
// This function calculates the last digit of X^N
// factorial
static int lastDigit( int X, int N)
{
// Initialize the last digit of N factorial as 1
int lastDigitNFactorial = 1;
// Calculate the last digit of N factorial
for ( int i = 2; i <= N; i++) {
lastDigitNFactorial
= (lastDigitNFactorial * (i % 10)) % 10;
}
// Calculate the last digit of X^(N factorial)
int lastDigitXToLastDigitNFactorial
= ( int )Math.Pow(X % 10, lastDigitNFactorial)
% 10;
// Return the result
return lastDigitXToLastDigitNFactorial;
}
static void Main( string [] args)
{
// Initialize X and N
int X = 18;
int N = 4;
// Print the result of lastDigit(X, N)
Console.WriteLine(lastDigit(X, N));
}
} |
// Javascript program implementation function last_digit(X, N) {
let last_digit_n_factorial = 1;
for (let i = 2; i <= N; i++) {
last_digit_n_factorial = (last_digit_n_factorial * (i % 10)) % 10;
}
let last_digit_X_to_last_digit_n_factorial = Math.pow(X % 10, last_digit_n_factorial) % 10;
return last_digit_X_to_last_digit_n_factorial;
} let X = 18; let N = 4; console.log(last_digit(X, N)); |
Output
6
Time complexity: O(N), where N is the input
Space complexity: O(1)