4 divisors code in python

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'''
Exercise 4
Create a program that asks the user for a number and then prints out a list of all the divisors of that number.
(If you don’t know what a divisor is, it is a number that divides evenly into another number. For example, 13 is
a divisor of 26 because 26 / 13 has no remainder.)
'''
num = int(raw_input("Enter a number: "))
listNum = list(range(1, num+1))
divisor = []
for each in listNum:
if num % each == 0:
divisor.append(each)
print(divisor)

Four Divisors, is a LeetCode problem from Math subdomain. In this post we will see how we can solve this challenge in Python

Problem Description

You can find the full details of the problem Four Divisors at LeetCode

Solution: Please check the main.py snippet for the solution.

This solution originally posted at: Github by @kamyu104


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    Given the number D, find the smallest number N such that it has exactly four divisors and the difference between any two of them is greater than or equal to D.

    Examples:

    Input: 1
    Output: 6
    Explanation: 6 has four divisors 1, 2, 3, and 6. 
    Difference between any two of them is always greater or equal to 1.

    Input: 2
    Output: 15
    Explanation: 15 has four divisors 1, 3, 5 and 15. 
    Difference between any two of them is always greater or equal to 2

    Input: 3
    Output: 55
    Explanation: 55 has four divisors 1, 5, 11 and 55. 
    Difference between any two of them is always greater than 3.

    Approach: It is obvious that ‘1’ and the number itself are the divisors of N. So the number which has exactly 4 divisors has its divisors as 1, a, b, a*b respectively. For the condition that it has exactly 4 divisors, both a and b must be prime. For the condition that the difference between any two of them should at least be D, start finding a from 1+d and check whether it is prime or not, If it is not prime then we will find the prime number just next to it. Similarly, start finding b from a + d and check whether it is prime or not, and do the same as done for finding a.

    Below is the implementation of the above approach.

    C++

    #include <bits/stdc++.h>

    using namespace std;

    int next_prime(int x)

    {

        if (x == 1 || x == 2) {

            return 2;

        }

        bool is_prime = false;

        while (!is_prime) {

            is_prime = true;

            for (int i = 2; i <= sqrt(x); i++) {

                if (x % i == 0) {

                    is_prime = false;

                }

            }

            x++;

        }

        return x - 1;

    }

    int findN(int D)

    {

        int a = 1 + D;

        a = next_prime(a);

        int b = a + D;

        b = next_prime(b);

        int N = a * b;

        return N;

    }

    int main()

    {

        int D = 2;

        int ans = findN(D);

        cout << ans;

    }

    Java

    import java.io.*;

    import java.lang.*;

    import java.util.*;

    class GFG {

      static int next_prime(int x)

      {

        if (x == 1 || x == 2) {

          return 2;

        }

        Boolean is_prime = false;

        while (!is_prime) {

          is_prime = true;

          for (int i = 2; i * i <= x; i++) {

            if (x % i == 0) {

              is_prime = false;

            }

          }

          x++;

        }

        return x - 1;

      }

      static int findN(int D)

      {

        int a = 1 + D;

        a = next_prime(a);

        int b = a + D;

        b = next_prime(b);

        int N = a * b;

        return N;

      }

      public static void main (String[] args) {

        int D = 2;

        int ans = findN(D);

        System.out.println(ans);

      }

    }

    Python3

    def next_prime(x):

        if (x == 1 or x == 2):

            return 2

        is_prime = False

        while (not is_prime):

            is_prime = True

            for i in range(2, int(x ** 0.5) + 1):

                if (x % i == 0):

                    is_prime = False

            x += 1

        return x - 1

    def findN(D):

        a = 1 + D

        a = next_prime(a)

        b = a + D

        b = next_prime(b)

        N = a * b

        return N

    D = 2

    ans = findN(D)

    print(ans)

    C#

    using System;

    class GFG {

      static int next_prime(int x)

      {

        if (x == 1 || x == 2) {

          return 2;

        }

        bool is_prime = false;

        while (!is_prime) {

          is_prime = true;

          for (int i = 2; i * i <= x; i++) {

            if (x % i == 0) {

              is_prime = false;

            }

          }

          x++;

        }

        return x - 1;

      }

      static int findN(int D)

      {

        int a = 1 + D;

        a = next_prime(a);

        int b = a + D;

        b = next_prime(b);

        int N = a * b;

        return N;

      }

      public static void Main () {

        int D = 2;

        int ans = findN(D);

        Console.WriteLine(ans);

      }

    }

    Javascript

    <script>

            function next_prime(x)

            {

                if (x == 1 || x == 2) {

                    return 2;

                }

                let is_prime = false;

                while (!is_prime) {

                    is_prime = true;

                    for (let i = 2; i <= Math.sqrt(x); i++) {

                        if (x % i == 0) {

                            is_prime = false;

                        }

                    }

                    x++;

                }

                return x - 1;

            }

            function findN(D)

            {

                let a = 1 + D;

                a = next_prime(a);

                let b = a + D;

                b = next_prime(b);

                let N = a * b;

                return N;

            }

            let D = 2;

            let ans = findN(D);

            document.write(ans);

        </script>

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


    How do you find divisors in Python?

    Finding divisors of a number with Python.
    def get_divisors(n): for i in range(1, int(n / 2) + 1): if n % i == 0: yield i yield n. ... .
    def prime_factors(n): i = 2 while i * i <= n: if n % i == 0: n /= i yield i else: i += 1 if n > 1: yield n..

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