# Program to find Nth term divisible by a or b

Given two integers and . The task is to find the Nth term which is divisible by either of or .

Examples :

```Input : a = 2, b = 5, N = 10
Output : 16

Input : a = 3, b = 7, N = 25
Output : 57
```

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

Naive Approach : A simple approach is to traverse over all the terms starting from 1 until we find the desired Nth term which is divisible by either of or . This solution has time complexity of O(N).

Efficient Approach : The idea is to use Binary search. Here we can calculate how many numbers from 1 to are divisible by either a or b by using formula: All the multiples of lcm(a, b) will be divisible by both and so we need to remove these terms. Now if the number of divisible terms is less than N we will increase the low position of binary search otherwise decrease high until number of divisible terms is equal to N.

Below is the implementation of the above idea :

## C++

 `// C++ program to find nth term ` `// divisible by a or b ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to return ` `// gcd of a and b ` `int` `gcd(``int` `a, ``int` `b) ` `{ ` `    ``if` `(a == 0) ` `        ``return` `b; ` ` `  `    ``return` `gcd(b % a, a); ` `} ` ` `  `// Function to calculate how many numbers ` `// from 1 to num are divisible by a or b ` `int` `divTermCount(``int` `a, ``int` `b, ``int` `lcm, ``int` `num) ` `{ ` `    ``// calculate number of terms divisible by a and ` `    ``// by b then, remove the terms which is are ` `    ``// divisible by both a and b ` `    ``return` `num / a + num / b - num / lcm; ` `} ` ` `  `// Binary search to find the nth term ` `// divisible by a or b ` `int` `findNthTerm(``int` `a, ``int` `b, ``int` `n) ` `{ ` `    ``// set low to 1 and high to max(a, b)*n, here ` `    ``// we have taken high as 10^18 ` `    ``int` `low = 1, high = INT_MAX, mid; ` `    ``int` `lcm = (a * b) / gcd(a, b); ` ` `  `    ``while` `(low < high) { ` `        ``mid = low + (high - low) / 2; ` ` `  `        ``// if the current term is less than ` `        ``// n then we need to increase low ` `        ``// to mid + 1 ` `        ``if` `(divTermCount(a, b, lcm, mid) < n) ` `            ``low = mid + 1; ` ` `  `        ``// if current term is greater than equal to ` `        ``// n then high = mid ` `        ``else` `            ``high = mid; ` `    ``} ` ` `  `    ``return` `low; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `a = 2, b = 5, n = 10; ` `    ``cout << findNthTerm(a, b, n) << endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find nth term  ` `// divisible by a or b  ` `class` `GFG ` `{ ` `// Function to return  ` `// gcd of a and b  ` `static` `int` `gcd(``int` `a, ``int` `b)  ` `{  ` `    ``if` `(a == ``0``)  ` `        ``return` `b;  ` ` `  `    ``return` `gcd(b % a, a);  ` `}  ` ` `  `// Function to calculate how many numbers  ` `// from 1 to num are divisible by a or b  ` `static` `int` `divTermCount(``int` `a, ``int` `b,  ` `                        ``int` `lcm, ``int` `num)  ` `{  ` `    ``// calculate number of terms  ` `    ``// divisible by a and by b then,  ` `    ``// remove the terms which is are  ` `    ``// divisible by both a and b  ` `    ``return` `num / a + num / b - num / lcm;  ` `}  ` ` `  `// Binary search to find the  ` `// nth term divisible by a or b  ` `static` `int` `findNthTerm(``int` `a, ``int` `b, ``int` `n)  ` `{  ` `    ``// set low to 1 and high to max(a, b)*n,   ` `    ``// here we have taken high as 10^18  ` `    ``int` `low = ``1``, high = Integer.MAX_VALUE, mid;  ` `    ``int` `lcm = (a * b) / gcd(a, b);  ` ` `  `    ``while` `(low < high)  ` `    ``{  ` `        ``mid = low + (high - low) / ``2``;  ` ` `  `        ``// if the current term is less  ` `        ``// than n then we need to increase ` `        ``// low to mid + 1  ` `        ``if` `(divTermCount(a, b, lcm, mid) < n)  ` `            ``low = mid + ``1``;  ` ` `  `        ``// if current term is greater  ` `        ``// than equal to n then high = mid  ` `        ``else` `            ``high = mid;  ` `    ``}  ` ` `  `    ``return` `low;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main (String[] args)  ` `{ ` `    ``int` `a = ``2``, b = ``5``, n = ``10``;  ` `    ``System.out.println(findNthTerm(a, b, n)); ` `} ` `}  ` ` `  `// This code is contributed by Smitha `

## Python3

 `# Python 3 program to find nth term  ` `# divisible by a or b  ` `import` `sys ` ` `  `# Function to return gcd of a and b  ` `def` `gcd(a, b): ` `    ``if` `a ``=``=` `0``: ` `        ``return` `b ` `    ``return` `gcd(b ``%` `a, a) ` ` `  `# Function to calculate how many numbers  ` `# from 1 to num are divisible by a or b  ` `def` `divTermCount(a, b, lcm, num): ` ` `  `    ``# calculate number of terms divisible  ` `    ``# by a and by b then, remove the terms  ` `    ``# which are divisible by both a and b  ` `    ``return` `num ``/``/` `a ``+` `num ``/``/` `b ``-` `num ``/``/` `lcm ` ` `  `# Binary search to find the nth term  ` `# divisible by a or b  ` `def` `findNthTerm(a, b, n): ` ` `  `    ``# set low to 1 and high to max(a, b)*n,  ` `    ``# here we have taken high as 10^18  ` `    ``low ``=` `1``; high ``=` `sys.maxsize ` `    ``lcm ``=` `(a ``*` `b) ``/``/` `gcd(a, b) ` `    ``while` `low < high: ` `        ``mid ``=` `low ``+` `(high ``-` `low) ``/``/` `2` ` `  `        ``# if the current term is less  ` `        ``# than n then we need to increase   ` `        ``# low to mid + 1  ` `        ``if` `divTermCount(a, b, lcm, mid) < n: ` `            ``low ``=` `mid ``+` `1` ` `  `        ``# if current term is greater  ` `        ``# than equal to n then high = mid  ` `        ``else``: ` `            ``high ``=` `mid ` `    ``return` `low ` ` `  `# Driver code ` `a ``=` `2``; b ``=` `5``; n ``=` `10` `print``(findNthTerm(a, b, n)) ` ` `  `# This code is contributed by Shrikant13 `

## C#

 `// C# program to find nth term  ` `// divisible by a or b  ` `using` `System; ` ` `  `class` `GFG ` `{ ` `// Function to return gcd of a and b  ` `static` `int` `gcd(``int` `a, ``int` `b)  ` `{  ` `    ``if` `(a == 0)  ` `        ``return` `b;  ` ` `  `    ``return` `gcd(b % a, a);  ` `}  ` ` `  `// Function to calculate how many numbers  ` `// from 1 to num are divisible by a or b  ` `static` `int` `divTermCount(``int` `a, ``int` `b,  ` `                        ``int` `lcm, ``int` `num)  ` `{  ` `    ``// calculate number of terms  ` `    ``// divisible by a and by b then,  ` `    ``// remove the terms which is are  ` `    ``// divisible by both a and b  ` `    ``return` `num / a + num / b - num / lcm;  ` `}  ` ` `  `// Binary search to find the  ` `// nth term divisible by a or b  ` `static` `int` `findNthTerm(``int` `a, ``int` `b, ``int` `n)  ` `{  ` `    ``// set low to 1 and high to max(a, b)*n,  ` `    ``// here we have taken high as 10^18  ` `    ``int` `low = 1, high = ``int``.MaxValue, mid;  ` `    ``int` `lcm = (a * b) / gcd(a, b);  ` ` `  `    ``while` `(low < high)  ` `    ``{  ` `        ``mid = low + (high - low) / 2;  ` ` `  `        ``// if the current term is less  ` `        ``// than n then we need to increase ` `        ``// low to mid + 1  ` `        ``if` `(divTermCount(a, b, lcm, mid) < n)  ` `            ``low = mid + 1;  ` ` `  `        ``// if current term is greater  ` `        ``// than equal to n then high = mid  ` `        ``else` `            ``high = mid;  ` `    ``}  ` ` `  `    ``return` `low;  ` `}  ` ` `  `// Driver code  ` `static` `public` `void` `Main () ` `{ ` `    ``int` `a = 2, b = 5, n = 10;  ` `    ``Console.WriteLine(findNthTerm(a, b, n)); ` `} ` `} ` ` `  `// This code is contributed by Sach_Code `

Output:

```16
```

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