Given two integers **X** and **Y** representing the total number of divisors and the number of composite divisors respectively, the task is to check if there exists an integer **N** which has exactly **X** divisors and **Y** are composite numbers.

**Examples:**

Input:X = 6, Y = 3Output:YESExplanation:

N = 18 is such a number.

The divisors of 18 are 1, 2, 3, 6, 9 and 18.

The composite divisors of 18 are 6, 9 and 18.Input:X = 7, Y = 3Output:NOExplanation:

We see that no such number exists that has 7 positive divisors out of which 3 are composite divisors.

**Approach:**

- Firstly calculate the number of prime divisors of a number, which is equal to:

** Number of prime divisors = Total number of divisors – Number of composite divisors – 1 **

- So, the number of prime divisors,
**C**=**X**–**Y – 1**

- Since every number has 1 as a factor and 1 is neither a prime number nor a composite number, we have to exclude it from being counted in the number of prime divisors.

- If the number of composite divisors is less than the number of prime divisors, then it is not possible to find such a number at all.

- So if the prime factorization of
**X**contains at least**C**distinct integers, then a solution is possible. Otherwise, we cannot find a number**N**that will satisfy the given conditions.

- Find the maximum number of values
**X**can be decomposed into such that each value is greater than**1**. In other words, we can find out the prime factorization of**X**. - If that prime factorization has a number of terms greater than or equal to
**C**, then such a number is possible.

Below is the implementation of the above approach:

## C++

`// C++ program to check if a number` `// exists having exactly X positive` `// divisors out of which Y are` `// composite divisors` `#include <bits/stdc++.h>` `using` `namespace` `std;` `int` `factorize(` `int` `N)` `{` ` ` `int` `count = 0;` ` ` `int` `cnt = 0;` ` ` `// Count the number of` ` ` `// times 2 divides N` ` ` `while` `((N % 2) == 0) {` ` ` `N = N / 2;` ` ` `count++;` ` ` `}` ` ` `cnt = cnt + count;` ` ` `// check for all possible` ` ` `// numbers that can divide it` ` ` `for` `(` `int` `i = 3; i <= ` `sqrt` `(N); i += 2) {` ` ` `count = 0;` ` ` `while` `(N % i == 0) {` ` ` `count++;` ` ` `N = N / i;` ` ` `}` ` ` `cnt = cnt + count;` ` ` `}` ` ` `// if N at the end` ` ` `// is a prime number.` ` ` `if` `(N > 2)` ` ` `cnt = cnt + 1;` ` ` `return` `cnt;` `}` `// Function to check if any` `// such number exists` `void` `ifNumberExists(` `int` `X, ` `int` `Y)` `{` ` ` `int` `C, dsum;` ` ` `C = X - Y - 1;` ` ` `dsum = factorize(X);` ` ` `if` `(dsum >= C)` ` ` `cout << ` `"YES \n"` `;` ` ` `else` ` ` `cout << ` `"NO \n"` `;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `X, Y;` ` ` `X = 6;` ` ` `Y = 4;` ` ` `ifNumberExists(X, Y);` ` ` `return` `0;` `}` |

## Java

`// Java program to check if a number` `// exists having exactly X positive` `// divisors out of which Y are` `// composite divisors` `import` `java.lang.Math;` `class` `GFG{` ` ` `public` `static` `int` `factorize(` `int` `N)` `{` ` ` `int` `count = ` `0` `;` ` ` `int` `cnt = ` `0` `;` ` ` ` ` `// Count the number of` ` ` `// times 2 divides N` ` ` `while` `((N % ` `2` `) == ` `0` `)` ` ` `{` ` ` `N = N / ` `2` `;` ` ` `count++;` ` ` `}` ` ` ` ` `cnt = cnt + count;` ` ` ` ` `// Check for all possible` ` ` `// numbers that can divide it` ` ` `for` `(` `int` `i = ` `3` `; i <= Math.sqrt(N); i += ` `2` `)` ` ` `{` ` ` `count = ` `0` `;` ` ` `while` `(N % i == ` `0` `)` ` ` `{` ` ` `count++;` ` ` `N = N / i;` ` ` `}` ` ` `cnt = cnt + count;` ` ` `}` ` ` ` ` `// If N at the end` ` ` `// is a prime number.` ` ` `if` `(N > ` `2` `)` ` ` `cnt = cnt + ` `1` `;` ` ` `return` `cnt;` `}` ` ` `// Function to check if any` `// such number exists` `public` `static` `void` `ifNumberExists(` `int` `X, ` `int` `Y)` `{` ` ` `int` `C, dsum;` ` ` `C = X - Y - ` `1` `;` ` ` `dsum = factorize(X);` ` ` ` ` `if` `(dsum >= C)` ` ` `System.out.println(` `"YES"` `);` ` ` `else` ` ` `System.out.println(` `"NO"` `);` `}` `// Driver code ` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `X, Y;` ` ` `X = ` `6` `;` ` ` `Y = ` `4` `;` ` ` ` ` `ifNumberExists(X, Y);` `}` `}` `// This code is contributed by divyeshrabadiya07` |

## Python3

`# Python3 program to check if a number exists` `# having exactly X positive divisors out of` `# which Y are composite divisors` `import` `math` `def` `factorize(N):` ` ` `count ` `=` `0` ` ` `cnt ` `=` `0` ` ` `# Count the number of` ` ` `# times 2 divides N` ` ` `while` `((N ` `%` `2` `) ` `=` `=` `0` `):` ` ` `N ` `=` `N ` `/` `/` `2` ` ` `count` `+` `=` `1` ` ` `cnt ` `=` `cnt ` `+` `count` ` ` `# Check for all possible` ` ` `# numbers that can divide it` ` ` `sq ` `=` `int` `(math.sqrt(N))` ` ` `for` `i ` `in` `range` `(` `3` `, sq, ` `2` `):` ` ` `count ` `=` `0` ` ` ` ` `while` `(N ` `%` `i ` `=` `=` `0` `):` ` ` `count ` `+` `=` `1` ` ` `N ` `=` `N ` `/` `/` `i` ` ` ` ` `cnt ` `=` `cnt ` `+` `count` ` ` `# If N at the end` ` ` `# is a prime number.` ` ` `if` `(N > ` `2` `):` ` ` `cnt ` `=` `cnt ` `+` `1` ` ` `return` `cnt` `# Function to check if any` `# such number exists` `def` `ifNumberExists(X, Y):` ` ` `C ` `=` `X ` `-` `Y ` `-` `1` ` ` `dsum ` `=` `factorize(X)` ` ` ` ` `if` `(dsum >` `=` `C):` ` ` `print` `(` `"YES"` `)` ` ` `else` `:` ` ` `print` `(` `"NO"` `)` `# Driver Code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` ` ` `X ` `=` `6` ` ` `Y ` `=` `4` ` ` `ifNumberExists(X, Y)` `# This code is contributed by chitranayal` |

## C#

`// C# program to check if a number` `// exists having exactly X positive` `// divisors out of which Y are` `// composite divisors` `using` `System;` `class` `GFG{` `public` `static` `int` `factorize(` `int` `N)` `{` ` ` `int` `count = 0;` ` ` `int` `cnt = 0;` ` ` ` ` `// Count the number of` ` ` `// times 2 divides N` ` ` `while` `((N % 2) == 0)` ` ` `{` ` ` `N = N / 2;` ` ` `count++;` ` ` `}` ` ` ` ` `cnt = cnt + count;` ` ` ` ` `// Check for all possible` ` ` `// numbers that can divide it` ` ` `for` `(` `int` `i = 3; i <= Math.Sqrt(N); i += 2)` ` ` `{` ` ` `count = 0;` ` ` `while` `(N % i == 0)` ` ` `{` ` ` `count++;` ` ` `N = N / i;` ` ` `}` ` ` `cnt = cnt + count;` ` ` `}` ` ` ` ` `// If N at the end` ` ` `// is a prime number.` ` ` `if` `(N > 2)` ` ` `cnt = cnt + 1;` ` ` `return` `cnt;` `}` ` ` `// Function to check if any` `// such number exists` `public` `static` `void` `ifNumberExists(` `int` `X, ` `int` `Y)` `{` ` ` `int` `C, dsum;` ` ` `C = X - Y - 1;` ` ` `dsum = factorize(X);` ` ` ` ` `if` `(dsum >= C)` ` ` `Console.WriteLine(` `"YES"` `);` ` ` `else` ` ` `Console.WriteLine(` `"NO"` `);` `}` `// Driver code` `public` `static` `void` `Main(` `string` `[] args)` `{` ` ` `int` `X, Y;` ` ` `X = 6;` ` ` `Y = 4;` ` ` ` ` `ifNumberExists(X, Y);` `}` `}` `// This code is contributed by AnkitRai01` |

## Javascript

`<script>` `// javascript program to check if a number` `// exists having exactly X positive` `// divisors out of which Y are` `// composite divisors` ` ` `function` `factorize(N) {` ` ` `var` `count = 0;` ` ` `var` `cnt = 0;` ` ` `// Count the number of` ` ` `// times 2 divides N` ` ` `while` `((N % 2) == 0) {` ` ` `N = N / 2;` ` ` `count++;` ` ` `}` ` ` `cnt = cnt + count;` ` ` `// Check for all possible` ` ` `// numbers that can divide it` ` ` `for` `(i = 3; i <= Math.sqrt(N); i += 2) {` ` ` `count = 0;` ` ` `while` `(N % i == 0) {` ` ` `count++;` ` ` `N = N / i;` ` ` `}` ` ` `cnt = cnt + count;` ` ` `}` ` ` `// If N at the end` ` ` `// is a prime number.` ` ` `if` `(N > 2)` ` ` `cnt = cnt + 1;` ` ` `return` `cnt;` ` ` `}` ` ` `// Function to check if any` ` ` `// such number exists` ` ` `function` `ifNumberExists(X , Y) {` ` ` `var` `C, dsum;` ` ` `C = X - Y - 1;` ` ` `dsum = factorize(X);` ` ` `if` `(dsum >= C)` ` ` `document.write(` `"YES"` `);` ` ` `else` ` ` `document.write(` `"NO"` `);` ` ` `}` ` ` `// Driver code` ` ` ` ` `var` `X, Y;` ` ` `X = 6;` ` ` `Y = 4;` ` ` `ifNumberExists(X, Y);` `// This code is contributed by todaysgaurav` `</script>` |

**Output:**

YES

**Time Complexity:** O (N^{ 1/2})**Auxiliary Space:** O (1)

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the **Essential Maths for CP Course** at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**