Count of all even numbers in the range [L, R] whose sum of digits is divisible by 3

Given two integers L and R. The task is to find the count of all even numbers in the range [L, R] whose sum of digits is divisible by 3.

Examples:

Input: L = 18, R = 36
Output: 4
18, 24, 30, 36 are the only numbers in the range [18, 36] which are even and whose sum of digits is divisible by 3.

Input: L = 7, R = 11
Output: 0
There is no number in the range [7, 11] which is even and whose sum of digits is divisible by 3.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive approach: Initialize count = 0 and for every number in the range [L, R], check if the number is divisible by 2 and sum of its digits is divisible by 3. If yes then increment the count. Print the count in the end.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to return the 
// sum of digits of x 
int sumOfDigits(int x) 
{ 
    int sum = 0; 
    while (x != 0) { 
        sum += x % 10; 
        x = x / 10; 
    } 
    return sum; 
} 
  
// Function to return the count 
// of required numbers 
int countNumbers(int l, int r) 
{ 
    int count = 0; 
    for (int i = l; i <= r; i++) { 
  
        // If i is divisible by 2 and 
        // sum of digits of i is divisible by 3 
        if (i % 2 == 0 && sumOfDigits(i) % 3 == 0) 
            count++; 
    } 
  
    // Return the required count 
    return count; 
} 
  
// Driver code 
int main() 
{ 
    int l = 1000, r = 6000; 
    cout << countNumbers(l, r); 
  
    return 0; 
} 

Java

// Java implementation of the approach  
class GFG 
{ 
  
// Function to return the  
// sum of digits of x  
static int sumOfDigits(int x)  
{  
    int sum = 0;  
    while (x != 0)  
    {  
        sum += x % 10;  
        x = x / 10;  
    }  
    return sum;  
}  
  
// Function to return the count  
// of required numbers  
static int countNumbers(int l, int r)  
{  
    int count = 0;  
    for (int i = l; i <= r; i++) 
    {  
  
        // If i is divisible by 2 and  
        // sum of digits of i is divisible by 3  
        if (i % 2 == 0 && sumOfDigits(i) % 3 == 0)  
            count++;  
    }  
  
    // Return the required count  
    return count;  
}  
  
// Driver code  
public static void main(String args[])  
{  
    int l = 1000, r = 6000;  
    System.out.println(countNumbers(l, r));  
}  
} 
  
// This code is contributed by Arnab Kundu 

Python3

# python implementation of the approach 
  
# Function to return the 
# sum of digits of x 
def sumOfDigits(x): 
    sum = 0
    while x != 0: 
        sum += x % 10
        x = x//10
    return sum
  
  
# Function to return the count 
# of required numbers 
def countNumbers(l, r): 
    count = 0
    for i in range(l, r + 1): 
  
        # If i is divisible by 2 and 
        # sum of digits of i is divisible by 3 
        if i % 2 == 0 and sumOfDigits(i) % 3 == 0: 
            count += 1
    return count 
  
# Driver code 
l = 1000; r = 6000
print(countNumbers(l, r)) 
  
# This code is contributed by Shrikant13 

C#

// C# implementation of the approach  
using System; 
  
class GFG 
{ 
  
// Function to return the  
// sum of digits of x  
static int sumOfDigits(int x)  
{  
    int sum = 0;  
    while (x != 0)  
    {  
        sum += x % 10;  
        x = x / 10;  
    }  
    return sum;  
}  
  
// Function to return the count  
// of required numbers  
static int countNumbers(int l, int r)  
{  
    int count = 0;  
    for (int i = l; i <= r; i++) 
    {  
  
        // If i is divisible by 2 and  
        // sum of digits of i is divisible by 3  
        if (i % 2 == 0 && sumOfDigits(i) % 3 == 0)  
            count++;  
    }  
  
    // Return the required count  
    return count;  
}  
  
// Driver code  
public static void Main()  
{  
    int l = 1000, r = 6000;  
    Console.WriteLine(countNumbers(l, r));  
}  
} 
  
// This code is contributed by Code_Mech. 

PHP

<?php 
// PHP implementation of the approach 
  
// Function to return the sum of 
// digits of x 
function sumOfDigits( $x) 
{ 
    $sum = 0; 
    while ($x != 0) 
    { 
        $sum += $x % 10; 
        $x = $x / 10; 
    } 
    return $sum; 
} 
  
// Function to return the count 
// of required numbers 
function countNumbers($l, $r) 
{ 
    $count = 0; 
    for ($i = $l; $i <= $r; $i++)  
    { 
  
        // If i is divisible by 2 and 
        // sum of digits of i is divisible by 3 
        if ($i % 2 == 0 &&  
            sumOfDigits($i) % 3 == 0) 
            $count++; 
    } 
  
    // Return the required count 
    return $count; 
} 
  
// Driver code 
$l = 1000; 
$r = 6000; 
echo countNumbers($l, $r); 
  
// This code is contributed by princiraj1992 
?> 

Output:

Time Complexity: O(r – l)

Efficient approach:

We have to check that the number is divisible by 2.
We have to check that the sum of digit is divisible by 3 which means that the number is divisible by 3.

So overall we have to check if a number is divisible by both 2 and 3, and since both 2 and 3 are co prime so we just have to check if a number is divisible by their product i.e. 6.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to return the count 
// of required numbers 
int countNumbers(int l, int r) 
{ 
  
    // Count of numbers in range 
    // which are divisible by 6 
    return ((r / 6) - (l - 1) / 6); 
} 
  
// Driver code 
int main() 
{ 
    int l = 1000, r = 6000; 
    cout << countNumbers(l, r); 
  
    return 0; 
} 

Java

// Java implementation of the approach 
class GFG 
{ 
  
// Function to return the count 
// of required numbers 
static int countNumbers(int l, int r) 
{ 
  
    // Count of numbers in range 
    // which are divisible by 6 
    return ((r / 6) - (l - 1) / 6); 
} 
  
// Driver code 
public static void main(String[] args) 
{ 
    int l = 1000, r = 6000; 
    System.out.println(countNumbers(l, r)); 
} 
} 
  
// This code is contributed by princiraj1992 

Python3

# Python3 implementation of the approach  
  
# Function to return the count  
# of required numbers  
def countNumbers(l, r) : 
  
    # Count of numbers in range  
    # which are divisible by 6  
    return ((r // 6) - (l - 1) // 6);  
  
# Driver code  
if __name__ == "__main__" : 
  
    l = 1000; r = 6000;  
    print(countNumbers(l, r));  
  
# This code is contributed by Ryuga 

C#

// C# implementation of the above approach  
using System;      
  
class GFG 
{ 
  
// Function to return the count 
// of required numbers 
static int countNumbers(int l, int r) 
{ 
  
    // Count of numbers in range 
    // which are divisible by 6 
    return ((r / 6) - (l - 1) / 6); 
} 
  
// Driver code 
public static void Main(String[] args) 
{ 
    int l = 1000, r = 6000; 
    Console.WriteLine(countNumbers(l, r)); 
} 
} 
  
// This code contributed by Rajput-Ji 

PHP

Output:

Time Complexity: O(1)

My Personal Notes

Suggest a Topic

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

PHP

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