# Check if the concatenation of first N natural numbers is divisible by 3

Given an integer N, the task is to check if the concatenation of first N natural numbers is divisible by 3 or not. Print Yes if divisible and No if not.

Examples:

Input: N = 3
Output: Yes
Explanation:
The concatenated number = 123
Since it is divisible by 3, the output is Yes
Input: N = 7
Output: No
Explanation: The concatenated number = 1234567
Since it is not divisible by 3, the output is No.

Naive Approach
The simplest approach is to concatenate the first N natural numbers and calculate the sum of digits of the resultant number and check if it is divisible by 3 or not.
Time Complexity: O(N)
Auxiliary Space: O(1)
Efficient Approach:
To optimize the above approach, we can observe a pattern. The concatenation of first N natural numbers is not divisible by 3 for the following series 1, 4, 7, 10, 13, 16, 19, and so on. The Nth term of the series is given by the formula 3×n +1. Hence, if (N – 1) is not divisible by 3, then the resultant number is divisible by 3, so print Yes. Otherwise, print No.
Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include `   `using` `namespace` `std;`   `// Function that returns True if` `// concatenation of first N natural` `// numbers is divisible by 3` `bool` `isDivisible(``int` `N)` `{` `    ``// Check using the formula` `    ``return` `(N - 1) % 3 != 0;` `}`     `// Driver Code` `int` `main()` `{` `    ``// Given Number` `    ``int` `N = 6;` `    `  `    ``// Function Call` `    ``if` `(isDivisible(N))` `        ``cout << (``"Yes"``);` `    `  `    ``else` `        ``cout << (``"No"``);` `    `  `    ``return` `0;` `}`   `// This code is contributed by Mohit Kumar`

## Java

 `// Java program for the above approach` `class` `GFG{` ` `  `// Function that returns True if` `// concatenation of first N natural` `// numbers is divisible by 3` `static` `boolean` `isDivisible(``int` `N)` `{` `    ``// Check using the formula` `    ``return` `(N - ``1``) % ``3` `!= ``0``;` `}` ` `  ` `  `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``// Given Number` `    ``int` `N = ``6``;` `     `  `    ``// Function Call` `    ``if` `(isDivisible(N))` `        ``System.out.println(``"Yes"``);` `     `  `    ``else` `        ``System.out.println(``"No"``);` `}` `}`   `// This code is contributed by Ritik Bansal`

## Python 3

 `# Python program for the above approach`   `# Function that returns True if` `# concatenation of first N natural` `# numbers is divisible by 3` `def` `isDivisible(N):`   `    ``# Check using the formula` `    ``return` `(N ``-` `1``) ``%` `3` `!``=` `0`     `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:`   `    ``# Given Number` `    ``N ``=` `6`   `    ``# Function Call` `    ``if` `(isDivisible(N)):` `        ``print``(``"Yes"``)`   `    ``else``:` `        ``print``(``"No"``)`

Output:

```Yes

```

Time Complexity: O(1)
Auxiliary Space: O(1)

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Improved By : mohit kumar 29, btc_148