# Smallest number having only 4 divisors with difference between any two at most D

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:1Output:6Explanation:6 has four divisors 1, 2, 3, and 6.

Difference between any two of them is always greater or equal to 1.

Input:2Output:15Explanation:15 has four divisors 1, 3, 5 and 15.

Difference between any two of them is always greater or equal to 2

Input:3Output:55Explanation: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++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `int` `next_prime(` `int` `x)` `{` ` ` `// Edge case` ` ` `if` `(x == 1 || x == 2) {` ` ` `return` `2;` ` ` `}` ` ` `// Checking if x is prime` ` ` `bool` `is_prime = ` `false` `;` ` ` `// Loop until next prime is found` ` ` `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;` `}` `// Function to find the number` `int` `findN(` `int` `D)` `{` ` ` `// Assuming 1+D as first required` ` ` `// divisor because it is` ` ` `// at distance D from 1` ` ` `int` `a = 1 + D;` ` ` `// Checking whether 1+D is prime or not` ` ` `// otherwise it will return prime number` ` ` `// just next to it` ` ` `a = next_prime(a);` ` ` `// Checking the same for next divisor` ` ` `int` `b = a + D;` ` ` `b = next_prime(b);` ` ` `int` `N = a * b;` ` ` `return` `N;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `D = 2;` ` ` `int` `ans = findN(D);` ` ` `cout << ans;` `}` |

## Java

`// Java program for the above approach` `import` `java.io.*;` `import` `java.lang.*;` `import` `java.util.*;` `class` `GFG {` ` ` `static` `int` `next_prime(` `int` `x)` ` ` `{` ` ` `// Edge case` ` ` `if` `(x == ` `1` `|| x == ` `2` `) {` ` ` `return` `2` `;` ` ` `}` ` ` `// Checking if x is prime` ` ` `Boolean is_prime = ` `false` `;` ` ` `// Loop until next prime is found` ` ` `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` `;` ` ` `}` ` ` `// Function to find the number` ` ` `static` `int` `findN(` `int` `D)` ` ` `{` ` ` `// Assuming 1+D as first required` ` ` `// divisor because it is` ` ` `// at distance D from 1` ` ` `int` `a = ` `1` `+ D;` ` ` `// Checking whether 1+D is prime or not` ` ` `// otherwise it will return prime number` ` ` `// just next to it` ` ` `a = next_prime(a);` ` ` `// Checking the same for next divisor` ` ` `int` `b = a + D;` ` ` `b = next_prime(b);` ` ` `int` `N = a * b;` ` ` `return` `N;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main (String[] args) {` ` ` `int` `D = ` `2` `;` ` ` `int` `ans = findN(D);` ` ` `System.out.println(ans);` ` ` `}` `}` `// This code is contributed by hrithikgarg03188.` |

## Python3

`# Python code for the above approach` `def` `next_prime(x):` ` ` ` ` `# Edge case` ` ` `if` `(x ` `=` `=` `1` `or` `x ` `=` `=` `2` `):` ` ` `return` `2` ` ` `# Checking if x is prime` ` ` `is_prime ` `=` `False` ` ` `# Loop until next prime is found` ` ` `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` `# Function to find the number` `def` `findN(D):` ` ` `# Assuming 1+D as first required` ` ` `# divisor because it is` ` ` `# at distance D from 1` ` ` `a ` `=` `1` `+` `D` ` ` `# Checking whether 1+D is prime or not` ` ` `# otherwise it will return prime number` ` ` `# just next to it` ` ` `a ` `=` `next_prime(a)` ` ` `# Checking the same for next divisor` ` ` `b ` `=` `a ` `+` `D` ` ` `b ` `=` `next_prime(b)` ` ` `N ` `=` `a ` `*` `b` ` ` `return` `N` `# Driver Code` `D ` `=` `2` `ans ` `=` `findN(D)` `print` `(ans)` `# This code is contributed by Saurabh Jaiswal` |

## C#

`// C# program for the above approach` `using` `System;` `class` `GFG {` ` ` `static` `int` `next_prime(` `int` `x)` ` ` `{` ` ` ` ` `// Edge case` ` ` `if` `(x == 1 || x == 2) {` ` ` `return` `2;` ` ` `}` ` ` `// Checking if x is prime` ` ` `bool` `is_prime = ` `false` `;` ` ` `// Loop until next prime is found` ` ` `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;` ` ` `}` ` ` `// Function to find the number` ` ` `static` `int` `findN(` `int` `D)` ` ` `{` ` ` ` ` `// Assuming 1+D as first required` ` ` `// divisor because it is` ` ` `// at distance D from 1` ` ` `int` `a = 1 + D;` ` ` `// Checking whether 1+D is prime or not` ` ` `// otherwise it will return prime number` ` ` `// just next to it` ` ` `a = next_prime(a);` ` ` `// Checking the same for next divisor` ` ` `int` `b = a + D;` ` ` `b = next_prime(b);` ` ` `int` `N = a * b;` ` ` `return` `N;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `Main () {` ` ` `int` `D = 2;` ` ` `int` `ans = findN(D);` ` ` `Console.WriteLine(ans);` ` ` `}` `}` `// This code is contributed by Samim Hossain Mondal.` |

## Javascript

`<script>` ` ` `// JavaScript code for the above approach ` ` ` `function` `next_prime(x)` ` ` `{` ` ` `// Edge case` ` ` `if` `(x == 1 || x == 2) {` ` ` `return` `2;` ` ` `}` ` ` `// Checking if x is prime` ` ` `let is_prime = ` `false` `;` ` ` `// Loop until next prime is found` ` ` `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 to find the number` ` ` `function` `findN(D)` ` ` `{` ` ` ` ` `// Assuming 1+D as first required` ` ` `// divisor because it is` ` ` `// at distance D from 1` ` ` `let a = 1 + D;` ` ` `// Checking whether 1+D is prime or not` ` ` `// otherwise it will return prime number` ` ` `// just next to it` ` ` `a = next_prime(a);` ` ` `// Checking the same for next divisor` ` ` `let b = a + D;` ` ` `b = next_prime(b);` ` ` `let N = a * b;` ` ` `return` `N;` ` ` `}` ` ` `// Driver Code` ` ` `let D = 2;` ` ` `let ans = findN(D);` ` ` `document.write(ans);` ` ` `// This code is contributed by Potta Lokesh` ` ` `</script>` |

**Output**

15

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