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Find the n-th number made of even digits only
  • Difficulty Level : Hard
  • Last Updated : 03 May, 2021

Given a number n, find out the n-th positive number made up of even digits (0, 2, 4, 6, 8) only. Firs few numbers made of even digits are 0, 2, 4, 6, 8, 20, 22, 24…….

Examples : 

Input  : 2
Output : 2
Second number made of 0, 2, 4, 6, 8 is 2

Input  : 10
Output : 28

Naive Approach 
A naive approach is to start from 0 and checking if it is made of only {0, 2, 4, 6, 8} and stop when you find n-th such number.  

C++




// Simple C++ program to find
// n-th number made of even
// digits only
#include<bits/stdc++.h>
using namespace std;
 
// function to calculate nth
// number made of even digits only
int findNthEvenDigitNumber(int n )
{
    // variable to note how
    // many such numbers have
    // been found till now
    int count = 0;
    for (int i = 0 ; ; i++)
    {
        int curr = i;
 
        // bool variable to check if
        // 1, 3, 5, 7, 9 is there or not
        bool isCurrEvenDigit = true ;
 
        // checking each digit
        // of the number
        while (curr != 0)
        {
            // If 1, 3, 5, 7, 9 is found
            // temp is changed to false
            if (curr % 10 == 1 || curr % 10 == 3 ||
                curr % 10 == 5 || curr % 10 == 7 ||
                curr % 10 == 9)
                isCurrEvenDigit = false;
            curr = curr / 10;
        }
 
        // temp is true it means that it
        // does not have 1, 3, 5, 7, 9
        if (isCurrEvenDigit == true)
            count++;
 
        // If nth such number is
        // found return it
        if (count == n)
            return i;
    }
}
 
// Driver Code
int main()
{
    cout << findNthEvenDigitNumber(2)
         << endl;
    cout << findNthEvenDigitNumber(10)
         << endl;
    return 0;
}

Java




// Simple Java program to
// find the n-th number made
// of even digits only
 
class GFG
{
    // function to calculate nth
    // number made of even digits only
    static int findNthEvenDigitNumber(int n )
    {
        // variable to note how
        // many such numbers have
        // been found till now
        int count = 0;
        for (int i = 0 ; ; i++)
        {
            int curr = i;
     
            // bool variable to check if
            // 1, 3, 5, 7, 9 is there or not
            boolean isCurrEvenDigit = true ;
     
            // checking each digit
            // of the number
            while (curr != 0)
            {
                // If 1, 3, 5, 7, 9 is found
                // temp is changed to false
                if (curr % 10 == 1 || curr % 10 == 3 ||
                    curr % 10 == 5 || curr % 10 == 7 ||
                    curr % 10 == 9)
                    isCurrEvenDigit = false;
                curr = curr / 10;
            }
     
            // temp is true it means that it
            // does not have 1, 3, 5, 7, 9
            if (isCurrEvenDigit == true)
                count++;
     
            // If nth such number
            // is found return it
            if (count == n)
                return i;
        }
    }
     
    // Driver Code
    public static void main (String[] args)
    {
        System.out.println(findNthEvenDigitNumber(2));
        System.out.println(findNthEvenDigitNumber(10));
    }
}

Python3




# Simple Python3 program to find nth
# number made of even digits only
 
# function to calculate nth number
# made of even digits only
def findNthEvenDigitNumber(n):
 
    # variable to note how many such
    # numbers have been found till now
    count = 0;
    i = 0;
    while (True):
     
        curr = i;
 
        # bool variable to check if
        # 1, 3, 5, 7, 9 is there or not
        isCurrEvenDigit = True;
 
        # checking each digit of the number
        while (curr != 0):
             
            # If 1, 3, 5, 7, 9 is found
            # temp is changed to false
            if (curr % 10 == 1 or curr % 10 == 3 or
                curr % 10 == 5 or curr % 10 == 7 or
                curr % 10 == 9):
                isCurrEvenDigit = False;
            curr = curr // 10;
 
        # temp is true it means that it
        # does not have 1, 3, 5, 7, 9
        if (isCurrEvenDigit == True):
            count += 1;
 
        # If nth such number is found,
        # return it
        if (count == n):
            return i;
         
        i += 1;
 
# Driver Code
print(findNthEvenDigitNumber(2));
print(findNthEvenDigitNumber(10));
 
# This code is contributed by mits

C#




// Simple C# program to
// find the n-th number
// made of even digits only
using System;
 
class GFG
{
    // function to calculate nth
    // number made of even digits only
    static int findNthEvenDigitNumber(int n )
    {
        // variable to note how
        // many such numbers have
        // been found till now
        int count = 0;
        for (int i = 0 ; ; i++)
        {
            int curr = i;
     
            // bool variable to check if
            // 1, 3, 5, 7, 9 is there or not
            bool isCurrEvenDigit = true ;
     
            // checking each digit
            // of the number
            while (curr != 0)
            {
                // If 1, 3, 5, 7, 9 is found
                // temp is changed to false
                if (curr % 10 == 1 || curr % 10 == 3 ||
                    curr % 10 == 5 || curr % 10 == 7 ||
                    curr % 10 == 9 )
                    isCurrEvenDigit = false;
                curr = curr / 10;
            }
     
            // temp is true it means that it
            // does not have 1, 3, 5, 7, 9
            if (isCurrEvenDigit == true)
                count++;
     
            // If nth such number
            // is found return it
            if (count == n)
                return i;
        }
    }
     
    // Driver code
    public static void Main ()
    {
        Console.WriteLine(findNthEvenDigitNumber(2));
        Console.WriteLine(findNthEvenDigitNumber(10));
    }
}
 
// This article is contributed by vt_m.

PHP




<?php
// Simple C++ program to find
// nth number made of even
// digits only
 
 
// function to calculate nth
// number made of even digits only
function findNthEvenDigitNumber($n )
{
    // variable to note how
    // many such numbers have
    // been found till now
    $count = 0;
    for ($i = 0 ; ; $i++)
    {
        $curr = $i;
 
        // bool variable to check if
        // 1, 3, 5, 7, 9 is there or not
        $isCurrEvenDigit = true ;
 
        // checking each digit
        // of the number
        while ($curr != 0)
        {
            // If 1, 3, 5, 7, 9 is found
            // temp is changed to false
            if ($curr % 10 == 1 || $curr % 10 == 3 ||
                $curr % 10 == 5 || $curr % 10 == 7 ||
                $curr % 10 == 9)
                $isCurrEvenDigit = false;
            $curr = $curr / 10;
        }
 
        // temp is true it means that it
        // does not have 1, 3, 5, 7, 9
        if ($isCurrEvenDigit == true)
            $count++;
 
        // If nth such number
        // is found return it
        if ($count == $n)
            return $i;
    }
}
 
// Driver Code
echo findNthEvenDigitNumber(2),"\n" ;
echo findNthEvenDigitNumber(10) ;
 
// This code is contributed by nitin mittal
?>

Javascript




<script>
 
// Simple JavaScript program to find
// n-th number made of even
// digits only
 
// Function to calculate nth
// number made of even digits only
function findNthEvenDigitNumber(n)
{
     
    // Variable to note how
    // many such numbers have
    // been found till now
    let count = 0;
     
    for(let i = 0;; i++)
    {
        let curr = i;
 
        // Bool variable to check if
        // 1, 3, 5, 7, 9 is there or not
        let isCurrEvenDigit = true;
 
        // Checking each digit
        // of the number
        while (curr != 0)
        {
             
            // If 1, 3, 5, 7, 9 is found
            // temp is changed to false
            if (curr % 10 == 1 || curr % 10 == 3 ||
                curr % 10 == 5 || curr % 10 == 7 ||
                curr % 10 == 9)
                isCurrEvenDigit = false;
                 
            curr = Math.floor(curr / 10);
        }
 
        // temp is true it means that it
        // does not have 1, 3, 5, 7, 9
        if (isCurrEvenDigit === true)
            count++;
 
        // If nth such number is
        // found return it
        if (count === n)
            return i;
    }
}
 
// Driver Code
document.write(findNthEvenDigitNumber(2) + "<br>");
document.write(findNthEvenDigitNumber(10) + "<br>");
 
// This code is contributed by Manoj.
 
</script>

Output : 

2
28

Efficient Approach 
We need to find numbers made of 5 digits, 0, 2, 4, 6 and 8. When we convert a number into base 5 number, it will only be made of numbers {0, 1, 2, 3, 4}. It can be clearly seen that each digit in the required digit set {0, 2, 4, 6, 8} is double the digit in the corresponding index of the base-5 digit set. So to find the n-th number made of only even digits follow the below mentioned steps
Step 1: Convert n to n-1 this is done so as to exclude zero. 
Step 2: Convert n to 5 base decimal number. 
Step 3: Multiply the above found number by 2. This is the required number



C++




// Efficient C++ program to
// find n-th number made of
// even digits only
#include<bits/stdc++.h>
using namespace std;
 
// function to find nth number
// made of even digits only
int findNthEvenDigitNumber(int n)
{
    // If n=1 return 0
    if (n == 1)
        return 0;
 
    // vector to store the digits
    // when converted into base 5
    vector< int> v;
 
    // Reduce n to n-1 to exclude 0
    n = n - 1;
  
    // Reduce n to base 5
    // number and store digits
    while (n > 0)
    {
        // pushing the digits
        // into vector
        v.push_back(n % 5);
        n = n / 5;
    }
 
    // variable to represent the
    // number after converting it
    // to base 5. Since the digits
    // are be in reverse order,
    // we traverse vector from back
    int result = 0;
    for (int i = v.size() - 1; i >= 0; i--)
    {
        result = result * 10;
        result = result + v[i];
    }
 
    // return 2*result (to convert
    // digits 0, 1, 2, 3, 4 to
    // 0, 2, 4, 6, 8.
    return 2*result;
}
 
// Driver Code
int main()
{
    cout << findNthEvenDigitNumber(2)
         << endl;
    cout << findNthEvenDigitNumber(10)
         << endl;
    return 0;
}

Java




import java.util.*;
// Efficient Java program to
// find n-th number made of
// even digits only
 
class GFG {
 
// function to find nth number
// made of even digits only
    static int findNthEvenDigitNumber(int n) {
        // If n=1 return 0
        if (n == 1) {
            return 0;
        }
 
        // vector to store the digits
        // when converted into base 5
        Vector< Integer> v = new Vector<>();
 
        // Reduce n to n-1 to exclude 0
        n = n - 1;
 
        // Reduce n to base 5
        // number and store digits
        while (n > 0) {
            // pushing the digits
            // into vector
            v.add(n % 5);
            n = n / 5;
        }
 
        // variable to represent the
        // number after converting it
        // to base 5. Since the digits
        // are be in reverse order,
        // we traverse vector from back
        int result = 0;
        for (int i = v.size() - 1; i >= 0; i--) {
            result = result * 10;
            result = result + v.get(i);
        }
 
        // return 2*result (to convert
        // digits 0, 1, 2, 3, 4 to
        // 0, 2, 4, 6, 8.
        return 2 * result;
    }
 
// Driver Code
    public static void main(String[] args) {
        System.out.println(findNthEvenDigitNumber(2));
        System.out.println(findNthEvenDigitNumber(10));
    }
}
// This code is contributed by PrinciRaj1992

Python3




# Efficient Python 3 program to find n-th
# number made of even digits only
 
# function to find nth number made of
# even digits only
def findNthEvenDigitNumber( n):
 
    # If n = 1 return 0
    if (n == 1):
        return 0
 
    # vector to store the digits
    # when converted into base 5
    v = []
 
    # Reduce n to n-1 to exclude 0
    n = n - 1
 
    # Reduce n to base 5 number and
    # store digits
    while (n > 0):
         
        # pushing the digits into vector
        v.append(n % 5)
        n = n // 5
 
    # variable to represent the number 
    # after converting it to base 5.
    # Since the digits are be in reverse
    # order, we traverse vector from back
    result = 0
    for i in range(len(v) - 1, -1, -1):
        result = result * 10
        result = result + v[i]
 
    # return 2*result (to convert
    # digits 0, 1, 2, 3, 4 to
    # 0, 2, 4, 6, 8.
    return 2 * result
 
# Driver Code
if __name__ == "__main__":
     
    print(findNthEvenDigitNumber(2))
    print(findNthEvenDigitNumber(10))
 
# This code is contributed by ita_c

C#




// Efficient C# program to
// find n-th number made of
// even digits only
using System;
using System.Collections;
 
class GFG {
 
    // function to find nth number
    // made of even digits only
    static int findNthEvenDigitNumber(int n)
    {
        // If n=1 return 0
        if (n == 1)
        {
            return 0;
        }
 
        // vector to store the digits
        // when converted into base 5
        ArrayList v = new ArrayList();
 
        // Reduce n to n-1 to exclude 0
        n = n - 1;
 
        // Reduce n to base 5
        // number and store digits
        while (n > 0)
        {
            // pushing the digits
            // into vector
            v.Add(n % 5);
            n = n / 5;
        }
 
        // variable to represent the
        // number after converting it
        // to base 5. Since the digits
        // are be in reverse order,
        // we traverse vector from back
        int result = 0;
        for (int i = v.Count - 1; i >= 0; i--)
        {
            result = result * 10;
            result = result + (int)v[i];
        }
 
        // return 2*result (to convert
        // digits 0, 1, 2, 3, 4 to
        // 0, 2, 4, 6, 8.
        return 2 * result;
    }
 
    // Driver Code
    public static void Main()
    {
        Console.WriteLine(findNthEvenDigitNumber(2));
        Console.WriteLine(findNthEvenDigitNumber(10));
    }
}
 
// This code is contributed by 29AjayKumar

PHP




<?php
// Efficient PHP program to find n-th
// number made of even digits only
 
// function to find nth number
// made of even digits only
function findNthEvenDigitNumber($n)
{
    // If n=1 return 0
    if ($n == 1)
        return 0;
 
    // vector to store the digits
    // when converted into base 5
    $v = array();
 
    // Reduce n to n-1 to exclude 0
    $n = $n - 1;
 
    // Reduce n to base 5
    // number and store digits
    while ($n > 0)
    {
        // pushing the digits
        // into vector
        array_push($v, $n % 5);
        $n = (int)($n / 5);
    }
 
    // variable to represent the number
    // after converting it to base 5.
    // Since the digits are be in
    // reverse order, we traverse vector
    // from back
    $result = 0;
    for ($i = count($v) - 1; $i >= 0; $i--)
    {
        $result = $result * 10;
        $result = $result + $v[$i];
    }
 
    // return 2*result (to convert
    // digits 0, 1, 2, 3, 4 to
    // 0, 2, 4, 6, 8.
    return 2 * $result;
}
 
// Driver Code
echo findNthEvenDigitNumber(2) . "\n";
echo findNthEvenDigitNumber(10) . "\n"
 
// This code is contributed by mits
?>

Javascript




<script>
// Efficient Javascript program to
// find n-th number made of
// even digits only
     
    // function to find nth number
    // made of even digits only
    function findNthEvenDigitNumber(n)
    {
        // If n=1 return 0
        if (n == 1) {
            return 0;
        }
  
        // vector to store the digits
        // when converted into base 5
        let v = [];
  
        // Reduce n to n-1 to exclude 0
        n = n - 1;
  
        // Reduce n to base 5
        // number and store digits
        while (n > 0) {
            // pushing the digits
            // into vector
            v.push(n % 5);
            n = Math.floor(n / 5);
        }
  
        // variable to represent the
        // number after converting it
        // to base 5. Since the digits
        // are be in reverse order,
        // we traverse vector from back
        let result = 0;
        for (let i = v.length - 1; i >= 0; i--) {
            result = result * 10;
            result = result + v[i];
        }
  
        // return 2*result (to convert
        // digits 0, 1, 2, 3, 4 to
        // 0, 2, 4, 6, 8.
        return 2 * result;
    }
     
    // Driver Code
    document.write(findNthEvenDigitNumber(2)+"<br>");
    document.write(findNthEvenDigitNumber(10));
     
    // This code is contributed by rag2127
</script>

Output : 

2
28

Time Complexity : O(log5(n))

This article is contributed by Ayush Jha. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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