# Find the number of integers x in range (1,N) for which x and x+1 have same number of divisors

Given an integer N. The task is to find the number of integers 1 < x < N, for which x and x + 1 have the same number of positive divisors.

Examples:

Input: N = 3
Output: 1
Divisors(1) = 1
Divisors(2) = 1 and 2
Divisors(3) = 1 and 3
Only valid x is 2.

Input: N = 15
Output: 2

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

Approach: Find the number of divisors of all numbers below N and store them in an array. And count the number of integers x such that x and x + 1 have the same number of positive divisors by running a loop.

Below is the implementation of the above approach:

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` `#define N 100005 ` ` `  `// To store number of divisors and ` `// Prefix sum of such numbers ` `int` `d[N], pre[N]; ` ` `  `// Function to find the number of integers ` `// 1 < x < N for which x and x + 1 have ` `// the same number of positive divisors ` `void` `Positive_Divisors() ` `{ ` `    ``// Count the number of divisors ` `    ``for` `(``int` `i = 1; i < N; i++) { ` ` `  `        ``// Run a loop upto sqrt(i) ` `        ``for` `(``int` `j = 1; j * j <= i; j++) { ` ` `  `            ``// If j is divisor of i ` `            ``if` `(i % j == 0) { ` ` `  `                ``// If it is perfect square ` `                ``if` `(j * j == i) ` `                    ``d[i]++; ` `                ``else` `                    ``d[i] += 2; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``int` `ans = 0; ` ` `  `    ``// x and x+1 have same number of ` `    ``// positive divisors ` `    ``for` `(``int` `i = 2; i < N; i++) { ` `        ``if` `(d[i] == d[i - 1]) ` `            ``ans++; ` `        ``pre[i] = ans; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``// Function call ` `    ``Positive_Divisors(); ` ` `  `    ``int` `n = 15; ` ` `  `    ``// Required answer ` `    ``cout << pre[n] << endl; ` ` `  `    ``return` `0; ` `} `

 `// Java implementation of the approach ` `class` `GFG ` `{ ` `     `  `static` `int` `N =``100005``; ` ` `  `// To store number of divisors and ` `// Prefix sum of such numbers ` `static` `int` `d[] = ``new` `int``[N], pre[] = ``new` `int``[N]; ` ` `  `// Function to find the number of integers ` `// 1 < x < N for which x and x + 1 have ` `// the same number of positive divisors ` `static` `void` `Positive_Divisors() ` `{ ` `    ``// Count the number of divisors ` `    ``for` `(``int` `i = ``1``; i < N; i++) ` `    ``{ ` ` `  `        ``// Run a loop upto sqrt(i) ` `        ``for` `(``int` `j = ``1``; j * j <= i; j++)  ` `        ``{ ` ` `  `            ``// If j is divisor of i ` `            ``if` `(i % j == ``0``) ` `            ``{ ` ` `  `                ``// If it is perfect square ` `                ``if` `(j * j == i) ` `                    ``d[i]++; ` `                ``else` `                    ``d[i] += ``2``; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``int` `ans = ``0``; ` ` `  `    ``// x and x+1 have same number of ` `    ``// positive divisors ` `    ``for` `(``int` `i = ``2``; i < N; i++) ` `    ``{ ` `        ``if` `(d[i] == d[i - ``1``]) ` `            ``ans++; ` `        ``pre[i] = ans; ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``// Function call ` `    ``Positive_Divisors(); ` ` `  `    ``int` `n = ``15``; ` ` `  `    ``// Required answer ` `    ``System.out.println(pre[n]); ` `} ` `} ` ` `  `/* This code contributed by PrinciRaj1992 */`

 `# Python3 implementation of the above approach  ` `from` `math ``import` `sqrt; ` ` `  `N ``=` `100005` ` `  `# To store number of divisors and  ` `# Prefix sum of such numbers  ` `d ``=` `[``0``] ``*` `N ` `pre ``=` `[``0``] ``*` `N ` ` `  `# Function to find the number of integers  ` `# 1 < x < N for which x and x + 1 have  ` `# the same number of positive divisors  ` `def` `Positive_Divisors() : ` `     `  `    ``# Count the number of divisors  ` `    ``for` `i ``in` `range``(N) : ` ` `  `        ``# Run a loop upto sqrt(i)  ` `        ``for` `j ``in` `range``(``1``, ``int``(sqrt(i)) ``+` `1``) : ` ` `  `            ``# If j is divisor of i  ` `            ``if` `(i ``%` `j ``=``=` `0``) : ` ` `  `                ``# If it is perfect square  ` `                ``if` `(j ``*` `j ``=``=` `i) : ` `                    ``d[i] ``+``=` `1` `                ``else` `: ` `                    ``d[i] ``+``=` `2` ` `  `    ``ans ``=` `0` ` `  `    ``# x and x+1 have same number of  ` `    ``# positive divisors  ` `    ``for` `i ``in` `range``(``2``, N) :  ` `        ``if` `(d[i] ``=``=` `d[i ``-` `1``]) : ` `            ``ans ``+``=` `1` `        ``pre[i] ``=` `ans ` `     `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``# Function call  ` `    ``Positive_Divisors() ` ` `  `    ``n ``=` `15` ` `  `    ``# Required answer  ` `    ``print``(pre[n])  ` ` `  `# This code is contributed by Ryuga `

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `static` `int` `N =100005; ` ` `  `// To store number of divisors and ` `// Prefix sum of such numbers ` `static` `int` `[]d = ``new` `int``[N];  ` `static` `int` `[]pre = ``new` `int``[N]; ` ` `  `// Function to find the number of integers ` `// 1 < x < N for which x and x + 1 have ` `// the same number of positive divisors ` `static` `void` `Positive_Divisors() ` `{ ` `    ``// Count the number of divisors ` `    ``for` `(``int` `i = 1; i < N; i++) ` `    ``{ ` ` `  `        ``// Run a loop upto sqrt(i) ` `        ``for` `(``int` `j = 1; j * j <= i; j++)  ` `        ``{ ` ` `  `            ``// If j is divisor of i ` `            ``if` `(i % j == 0) ` `            ``{ ` ` `  `                ``// If it is perfect square ` `                ``if` `(j * j == i) ` `                    ``d[i]++; ` `                ``else` `                    ``d[i] += 2; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``int` `ans = 0; ` ` `  `    ``// x and x+1 have same number of ` `    ``// positive divisors ` `    ``for` `(``int` `i = 2; i < N; i++) ` `    ``{ ` `        ``if` `(d[i] == d[i - 1]) ` `            ``ans++; ` `        ``pre[i] = ans; ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``// Function call ` `    ``Positive_Divisors(); ` ` `  `    ``int` `n = 15; ` ` `  `    ``// Required answer ` `    ``Console.WriteLine(pre[n]); ` `} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

 ` `

Output:
```2
```

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