Check whether a given number is Polydivisible or Not

Given an integer n, find whether n is a Polydivisible or not. In mathematics, a number is called Polydivisible if it follows some unique properties. The number should not have any leading zeroes. The number formed by first i digits of the input number should be divisible by i, where . If any number follow these properties then it is called Polydivisible number.
Examples:

Input: 345654
Output: 345654 is Polydivisible number.
Explanation:
The first digit of the number is non-zero.
The number formed by the first 2 digits(34)
is divisible by 2. The number formed by the
first 3 digits(345) is divisible by 3.
The number formed by the first 4 digits(3456)
is divisible by 4. The number formed by the
first 5 digits(34565) is divisible by 5.
The number formed by the first 6 digits(345654)
is divisible by 6.

Input: 130
Output: 130 is Not Polydivisible number.

Input: 129
Output: 129 is Polydivisible number.

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

Approach: The idea is very simple.

1. Extract all the digits of the array and store them in an array.
2. Pick first 2 digits and form a number and check if it is divisible by 2.
3. Pick ith digit and append to the existing number and check if the number is divisible by i.
4. If all the above conditions are satisfied until all the digits are exhausted,then the given number is Polydivisible.

Below is the implementation of the above approach.

C++

 // CPP program to check whether // a number is polydivisible or not #include using namespace std;    // function to check polydivisible // number void check_polydivisible(int n) {     int N = n;     vector digit;        // digit extraction of input number     while (n > 0) {            // store the digits in an array         digit.push_back(n % 10);         n /= 10;     }     reverse(digit.begin(), digit.end());        bool flag = true;     n = digit;     for (int i = 1; i < digit.size(); i++) {            // n contains first i digits         n = n * 10 + digit[i];            // n should be divisible by i         if (n % (i + 1) != 0) {             flag = false;             break;         }     }     if (flag)         cout << N << " is Polydivisible number.";     else         cout << N << " is Not Polydivisible number."; }    int main() {     int n = 345654;     check_polydivisible(n); }

Java

 // Java program to check whether // a number is polydivisible or not import java.util.*; import java.io.*;    class GFG {            // function to check polydivisible     // number     static void check_polydivisible(int n)     {         int N = n;         Vector digit = new Vector();                // digit extraction of input number         while (n > 0) {                    // store the digits in an array             digit.add(new Integer(n % 10));             n /= 10;         }         Collections.reverse(digit);                boolean flag = true;         n = digit.get(0);         for (int i = 1; i < digit.size(); i++) {                    // n contains first i digits             n = n * 10 + digit.get(i);                    // n should be divisible by i             if (n % (i + 1) != 0) {                 flag = false;                 break;             }         }         if (flag)             System.out.println(N + " is Polydivisible number.");         else             System.out.println(N + " is Not Polydivisible number.");     }            // Driver code     public static void main (String[] args)      {         int n = 345654;         check_polydivisible(n);     } }

Python3

 # Python 3 program to check whether # a number is polydivisible or not    # function to check polydivisible # number def check_polydivisible(n):     N = n     digit = []        # digit extraction of input number     while (n > 0):                    # store the digits in an array         digit.append(n % 10)         n /= 10        digit.reverse()        flag = True     n = digit     for i in range(1, len(digit), 1):                    # n contains first i digits         n = n * 10 + digit[i]            # n should be divisible by i         if (n % (i + 1) != 0):             flag = False             break            if (flag):         print(N, "is Polydivisible number.");     else:         print(N, "is Not Polydivisible number.")    # Driver Code if __name__ == '__main__':     n = 345654     check_polydivisible(n)           # This code is contributed by # Sahil_Shelangia

Output:

345654 is Polydivisible number.

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