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

  • Difficulty Level : Easy
  • Last Updated : 25 Mar, 2021

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:
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:
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++




// 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
?>

Javascript




<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>
Output: 



834

 

Time Complexity: O(r – l)
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++




// 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




<?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.
?>

Javascript




<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>
Output: 
834

 

Time Complexity: O(1)
 

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