# 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 ` `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

 ` `

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 ` `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:

```834
```

Time Complexity: O(1)

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