# Number of elements with even factors in the given range

Given a range [n, m], the task is to find the number of elements that have even number of factors in the given range (n and m inclusive).

**Examples :**

Input:n = 5, m = 20Output:14 The numbers with even factors are 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20.Input:n = 5, m = 100Output: 88

A **Simple Solution** is to loop through all numbers starting from n. For every number, check if it has an even number of factors. If it has an even number of factors then increment count of such numbers and finally print the number of such elements. To find all divisors of a natural number efficiently, refer All divisors of a natural number

An **Efficient Solution** is to find the numbers with odd number of factors i.e only the perfect squares have odd number of factors, so all numbers other than perfect squares will have even number of factors. So, find the count of perfect squares in the range and subtract from the total numbers i.e. **m-n+1** .

Below is the implementation of the above approach:

## C++

`// C++ implementation of the above approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to count the perfect squares ` `int` `countOddSquares(` `int` `n, ` `int` `m) ` `{ ` ` ` `return` `(` `int` `)` `pow` `(m, 0.5) - (` `int` `)` `pow` `(n - 1, 0.5); ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `n = 5, m = 100; ` ` ` `cout << ` `"Count is "` ` ` `<< (m - n + 1) - countOddSquares(n, m); ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## Java

`// Java implementation of the above approach ` `import` `java.io.*; ` ` ` `class` `GFG ` `{ ` ` ` `// Function to count the perfect squares ` `static` `int` `countOddSquares(` `int` `n, ` `int` `m) ` `{ ` ` ` `return` `(` `int` `)Math.pow(m, ` `0.5` `) - ` ` ` `(` `int` `)Math.pow(n - ` `1` `, ` `0.5` `); ` `} ` ` ` `// Driver code ` `public` `static` `void` `main (String[] args) ` `{ ` ` ` `int` `n = ` `5` `, m = ` `100` `; ` ` ` `System.out.println(` `"Count is "` `+ ((m - n + ` `1` `) ` ` ` `- countOddSquares(n, m))); ` `} ` `} ` ` ` `// This code is contributed by ajit.. ` |

*chevron_right*

*filter_none*

## Python 3

`# Python3 implementation of the ` `# above approach ` ` ` `# Function to count the perfect squares ` `def` `countOddSquares(n, m) : ` ` ` `return` `(` `int` `(` `pow` `(m, ` `0.5` `)) ` `-` ` ` `int` `(` `pow` `(n ` `-` `1` `, ` `0.5` `))) ` ` ` `# Driver code ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` ` ` `n ` `=` `5` `; m ` `=` `100` `; ` ` ` `print` `(` `"Count is"` `, (m ` `-` `n ` `+` `1` `) ` `-` ` ` `countOddSquares(n, m)) ` ` ` `# This code is contributed by Ryuga ` |

*chevron_right*

*filter_none*

## C#

`// C# implementation of the above approach ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `// Function to count the perfect squares ` `static` `int` `countOddSquares(` `int` `n, ` `int` `m) ` `{ ` ` ` `return` `(` `int` `)Math.Pow(m, 0.5) - ` ` ` `(` `int` `)Math.Pow(n - 1, 0.5); ` `} ` ` ` `// Driver code ` `static` `public` `void` `Main () ` `{ ` ` ` `int` `n = 5, m = 100; ` ` ` `Console.WriteLine(` `"Count is "` `+ ((m - n + 1) ` ` ` `- countOddSquares(n, m))); ` `} ` `} ` ` ` `// This Code is contributed by akt_mit. ` |

*chevron_right*

*filter_none*

## PHP

`<?php ` `// PHP implementation of the ` `// above approach ` ` ` `// Function to count the perfect squares ` `function` `countOddSquares(` `$n` `, ` `$m` `) ` `{ ` ` ` `return` `(int)pow(` `$m` `, 0.5) - ` ` ` `(int)pow(` `$n` `- 1, 0.5); ` `} ` ` ` `// Driver code ` `$n` `= 5; ` `$m` `= 100; ` `echo` `"Count is "` `, (` `$m` `- ` `$n` `+ 1) - ` ` ` `countOddSquares(` `$n` `, ` `$m` `); ` ` ` `// This code is contributed by ajit ` `?> ` |

*chevron_right*

*filter_none*

**Output:**

Count is 88

## Recommended Posts:

- Number of elements with odd factors in given range
- Find number of factors of N when location of its two factors whose product is N is given
- Sum of all even factors of numbers in the range [l, r]
- Sum of all odd factors of numbers in the range [l, r]
- K-Primes (Numbers with k prime factors) in a range
- Count numbers from range whose prime factors are only 2 and 3
- Count elements in the given range which have maximum number of divisors
- Count numbers in a range having GCD of powers of prime factors equal to 1
- Prime factors of LCM of array elements
- Minimum elements to be added in a range so that count of elements is divisible by K
- Queries on sum of odd number digit sums of all the factors of a number
- Number with maximum number of prime factors
- Sum of all the factors of a number
- Prime factors of a big number
- Product of factors of number

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.