Program to find the last digit of X in base Y

Given a positive integer X and Y, the task is to find the last digit of X in the given base Y.

Examples:

Input: X = 10, Y = 7
Output: 3
10 is 13 in base 9 with last digit 3

Input: X = 55, Y = 3
Output: 1
55 is 3 in base 601 with last digit 1

Approach:



  • When we try to convert X into the base Y
  • We repeatedly divide X by base Y and store the remainder.
  • So the final result comprises of the remainders in the order of division steps.
  • Lets say the remainder of division step 1 is p, step 2 is q, step 3 is r
  • Then the resultant number in base Y will be rqp
  • And the last digit will be p
  • Therefore, we just need to find the first remainder of X when divided by Y to get the lsat digit in X in base Y.
    last digit = X % Y
    

Below is the implementation of the above approach:

C++

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// C++ Program to find
// the last digit of X in base Y
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the last
// digit of X in base Y
void last_digit(int X, int Y)
{
    cout << X % Y;
}
  
// Driver code
int main()
{
    int X = 55, Y = 3;
    last_digit(X, Y);
    return 0;
}

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Java

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// Java Program to find
// the last digit of X in base Y
class GFG 
{
  
// Function to find the last
// digit of X in base Y
static void last_digit(int X, int Y)
{
    System.out.print(X % Y);
}
  
// Driver code
public static void main(String []args)
{
    int X = 55, Y = 3;
    last_digit(X, Y);
}
}
  
// This code is contributed by Rajput-Ji

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Python3

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# Python3 Program to find 
# the last digit of X in base Y 
  
# Function to find the last 
# digit of X in base Y 
def last_digit(X, Y) :
  
    print(X % Y); 
  
# Driver code 
if __name__ == "__main__" :
  
    X = 55; Y = 3
    last_digit(X, Y);
  
# This code is contributed 
# by AnkitRai01

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C#

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// C# Program to find the last digit
// of X in base Y
using System;
  
class GFG 
{
  
// Function to find the last
// digit of X in base Y
static void last_digit(int X, int Y)
{
    Console.Write(X % Y);
}
  
// Driver code
public static void Main(String []args)
{
    int X = 55, Y = 3;
    last_digit(X, Y);
}
}
  
// This code is contributed by Rajput-Ji

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Output:

1

Time Complexity: O(1)




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Improved By : AnkitRai01, Rajput-Ji