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. 
 

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++ 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 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

# 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# 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 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
?>

<script>
// JavaScript implementation of the approach

// Function to return the
// sum of digits of x
function sumOfDigits(x)
{
    let sum = 0;
    while (x != 0) {
        sum += x % 10;
        x = Math.floor(x / 10);
    }
    return sum;
}

// Function to return the count
// of required numbers
function countNumbers(l, r)
{
    let count = 0;
    for (let 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
    let l = 1000, r = 6000;
    document.write(countNumbers(l, r));

// This code is contributed by Manoj.
</script>

834

Time Complexity: O(r - l), as we are traversing from l to r.

Auxiliary Space: O(1), as we are not using any extra space.
Efficient approach: 
 

1. We have to check that the number is divisible by 2.
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++ 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 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 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# 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
// PHP implementation of the approach

// Function to return the count
// of required numbers
function countNumbers($l, $r)
{

    // Count of numbers in range
    // which are divisible by 6
    return ((int)($r / 6) - 
            (int)(($l - 1) / 6));
}

// Driver code
$l = 1000; $r = 6000;
echo(countNumbers($l, $r));

// This code is contributed 
// by Code_Mech.
?>

<script>

// Javascript implementation of the approach

// Function to return the count
// of required numbers
function countNumbers(l, r)
{

    // Count of numbers in range
    // which are divisible by 6
    return (parseInt(r / 6) - parseInt((l - 1) / 6));
}

// Driver code
var l = 1000, r = 6000;
document.write(countNumbers(l, r));

</script>

834

Time Complexity: O(1), as we are not traversing or using any loops.
 Auxiliary Space: O(1), as we are not using any extra space.