Largest Even and Odd N-digit numbers of base B

Given an integer N and base B, the task is to find the largest Even and Odd N-digit numbers of Base B in decimal form.

Examples:

Input: N = 2, B = 5
Output:
Even = 24
Odd = 23
Explanation:
Largest Even Number of 2 digits in base 5 = 44 which is 24 in decimal form.
Largest Odd Number of 2 digits in base 5 = 43 which is 23 in decimal form.

Input: N = 2, B = 10
Output:
Even = 98
Odd = 99

Approach:
To get the largest Even and Odd N-digits number of base B in decimal form is given by:



  1. If Base B is Even, then:
    • Largest N-digit even number is (BN – 2).
    • Largest N-digit odd number is (BN – 1).
  2. If Base B is Odd, then:
    • Largest N-digit Even number is (BN – 1).
    • Largest N-digit Odd number is (BN – 2).

Below is the implementation of the above approach:

C++

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// C++ implementation of the
// above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to print the largest
// N-digit even and odd numbers
// of base B
void findNumbers(int n, int b)
{
    // Intialise the Number
    int even = 0, odd = 0;
  
    // If Base B is even, then
    // B^n will give largest
    // Even number of N+1 digit
    if (b % 2 == 0) {
  
        // To get even number of
        // N digit subtract 2 from
        // B^n
        even = pow(b, n) - 2;
  
        // To get odd number of
        // N digit subtract 1 from
        // B^n
        odd = pow(b, n) - 1;
    }
  
    // If Base B is odd, then
    // B^n will give largest
    // Odd number of N+1 digit
    else {
  
        // To get even number of
        // N digit subtract 1 from
        // B^n
        even = pow(b, n) - 1;
  
        // To get odd number of
        // N digit subtract 2 from
        // B^n
        odd = pow(b, n) - 2;
    }
    cout << "Even Number = " << even << '\n';
    cout << "Odd Number = " << odd;
}
  
// Driver's Code
int main()
{
    int N = 2, B = 5;
  
    // Function to find the
    // numbers
    findNumbers(N, B);
    return 0;
}

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Java

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// Java implementation of the
// above approach
import java.util.*;
  
class GFG{
   
// Function to print the largest
// N-digit even and odd numbers
// of base B
static void findNumbers(int n, int b)
{
    // Intialise the Number
    double even = 0, odd = 0;
   
    // If Base B is even, then
    // B^n will give largest
    // Even number of N+1 digit
    if (b % 2 == 0) {
   
        // To get even number of
        // N digit subtract 2 from
        // B^n
        even = Math.pow(b, n) - 2;
   
        // To get odd number of
        // N digit subtract 1 from
        // B^n
        odd = Math.pow(b, n) - 1;
    }
   
    // If Base B is odd, then
    // B^n will give largest
    // Odd number of N+1 digit
    else {
   
        // To get even number of
        // N digit subtract 1 from
        // B^n
        even = Math.pow(b, n) - 1;
   
        // To get odd number of
        // N digit subtract 2 from
        // B^n
        odd = Math.pow(b, n) - 2;
    }
    System.out.println("Even Number = " +  (int)even );
    System.out.print("Odd Number = " +  (int)odd);
}
   
// Driver's Code
public static void main(String[] args)
{
    int N = 2, B = 5;
   
    // Function to find the
    // numbers
    findNumbers(N, B);
}
}
  
// This code is contributed by Rajput-Ji

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Python3

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# Python implementation of the
# above approach
  
# Function to print the largest
# N-digit even and odd numbers
# of base B
def findNumbers(n, b):
    # Intialise the Number
    even = 0;
    odd = 0;
  
    # If Base B is even, then
    # B^n will give largest
    # Even number of N+1 digit
    if (b % 2 == 0):
  
        # To get even number of
        # N digit subtract 2 from
        # B^n
        even = pow(b, n) - 2;
  
        # To get odd number of
        # N digit subtract 1 from
        # B^n
        odd = pow(b, n) - 1;
      
  
    # If Base B is odd, then
    # B^n will give largest
    # Odd number of N+1 digit
    else:
  
        # To get even number of
        # N digit subtract 1 from
        # B^n
        even = pow(b, n) - 1;
  
        # To get odd number of
        # N digit subtract 2 from
        # B^n
        odd = pow(b, n) - 2;
      
    print("Even Number = ",int(even));
    print("Odd Number = ", int(odd));
  
# Driver's Code
if __name__ == '__main__':
    N = 2;
    B = 5;
  
    # Function to find the
    # numbers
    findNumbers(N, B);
      
# This code is contributed by 29AjayKumar

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C#

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// C# implementation of the
// above approach
using System;
  
class GFG{
    
// Function to print the largest
// N-digit even and odd numbers
// of base B
static void findNumbers(int n, int b)
{
    // Intialise the Number
    double even = 0, odd = 0;
    
    // If Base B is even, then
    // B^n will give largest
    // Even number of N+1 digit
    if (b % 2 == 0) {
    
        // To get even number of
        // N digit subtract 2 from
        // B^n
        even = Math.Pow(b, n) - 2;
    
        // To get odd number of
        // N digit subtract 1 from
        // B^n
        odd = Math.Pow(b, n) - 1;
    }
    
    // If Base B is odd, then
    // B^n will give largest
    // Odd number of N+1 digit
    else {
    
        // To get even number of
        // N digit subtract 1 from
        // B^n
        even = Math.Pow(b, n) - 1;
    
        // To get odd number of
        // N digit subtract 2 from
        // B^n
        odd = Math.Pow(b, n) - 2;
    }
    Console.WriteLine("Even Number = " +  (int)even );
    Console.Write("Odd Number = " +  (int)odd);
}
    
// Driver's Code
public static void Main(String[] args)
{
    int N = 2, B = 5;
    
    // Function to find the
    // numbers
    findNumbers(N, B);
}
}
  
// This code is contributed by 29AjayKumar

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Output:

Even Number = 24
Odd Number = 23

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