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# Find the Nth term divisible by a or b or c

• Difficulty Level : Medium
• Last Updated : 10 May, 2021

Given four integers a, b, c, and N. The task is to find the Nth term which is divisible by either of a, b or c.
Examples:

Input: a = 2, b = 3, c = 5, N = 10
Output: 14
Sequence is 2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 15, 16…
Input: a = 3, b = 5, c = 7, N = 10
Output: 18

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 a, b or c. 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 num are divisible by either a, b or c by using the formula: (num / a) + (num / b) + (num / c) – (num / lcm(a, b)) – (num / lcm(b, c)) – (num / lcm(a, c)) + (num / lcm(a, b, c))
The above formula is derived using set theory
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#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 return the lcm of a and b``int` `lcm(``int` `a, ``int` `b)``{``    ``return` `(a * b) / gcd(a, b);``}` `// Function to return the count of numbers``// from 1 to num which are divisible by a, b or c``int` `divTermCount(``int` `a, ``int` `b, ``int` `c, ``int` `num)``{` `    ``// Calculate number of terms divisible by a and``    ``// by b and by c then, remove the terms which is are``    ``// divisible by both a and b, both b and c, both``    ``// c and a and then add which are divisible by a and``    ``// b and c``    ``return` `((num / a) + (num / b) + (num / c)``            ``- (num / lcm(a, b))``            ``- (num / lcm(b, c))``            ``- (num / lcm(a, c))``            ``+ (num / lcm(a, lcm(b, c))));``}` `// Function to find the nth term``// divisible by a, b or c``// by using binary search``int` `findNthTerm(``int` `a, ``int` `b, ``int` `c, ``int` `n)``{` `    ``// Set low to 1 and high to max (a, b, c) * n``    ``int` `low = 1, high = INT_MAX, mid;` `    ``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, c, 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 = 3, c = 5, n = 10;` `    ``cout << findNthTerm(a, b, c, n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``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 return the lcm of a and b``    ``static` `int` `lcm(``int` `a, ``int` `b)``    ``{``        ``return` `(a * b) / gcd(a, b);``    ``}` `    ``// Function to return the count of numbers``    ``// from 1 to num which are divisible by a, b or c``    ``static` `int` `divTermCount(``int` `a, ``int` `b, ``int` `c, ``int` `num)``    ``{` `        ``// Calculate number of terms divisible by a and``        ``// by b and by c then, remove the terms which is are``        ``// divisible by both a and b, both b and c, both``        ``// c and a and then add which are divisible by a and``        ``// b and c``        ``return` `((num / a) + (num / b) + (num / c)``                ``- (num / lcm(a, b))``                ``- (num / lcm(b, c))``                ``- (num / lcm(a, c))``                ``+ (num / lcm(a, lcm(b, c))));``    ``}` `    ``// Function to find the nth term``    ``// divisible by a, b or c``    ``// by using binary search``    ``static` `int` `findNthTerm(``int` `a, ``int` `b, ``int` `c, ``int` `n)``    ``{` `        ``// Set low to 1 and high to max (a, b, c) * n``        ``int` `low = ``1``, high = Integer.MAX_VALUE, mid;` `        ``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, c, 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 = ``3``, c = ``5``, n = ``10``;` `        ``System.out.println(findNthTerm(a, b, c, n));` `    ``}``}` `// This code is contributed by``// Rajnis09`

## Python

 `# Python3 implementation of the approach` `# Function to return``# gcd of a and b``def` `gcd(a, b):` `    ``if` `(a ``=``=` `0``):``        ``return` `b` `    ``return` `gcd(b ``%` `a, a)` `# Function to return the lcm of a and b``def` `lcm(a, b):` `    ``return` `((a ``*` `b) ``/``/` `gcd(a, b))`  `# Function to return the count of numbers``# from 1 to num which are divisible by a, b or c``def` `divTermCount(a, b, c, num):` `    ``# Calculate number of terms divisible by a and``    ``# by b and by c then, remove the terms which is are``    ``# divisible by both a and b, both b and c, both``    ``# c and a and then add which are divisible by a and``    ``# b and c``    ``return` `((num ``/``/` `a) ``+` `(num ``/``/` `b) ``+` `(num ``/``/` `c)``            ``-` `(num ``/``/` `lcm(a, b))``            ``-` `(num ``/``/` `lcm(b, c))``            ``-` `(num ``/``/` `lcm(a, c))``            ``+` `(num ``/``/` `lcm(a, lcm(b, c))))`  `# Function to find the nth term``# divisible by a, b or c``# by using binary search``def` `findNthTerm(a, b, c, n):`  `    ``# Set low to 1 and high to max (a, b, c) * n``    ``low ``=` `1``    ``high ``=` `10``*``*``9``    ``mid``=``0` `    ``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, c, 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 ``=` `3``c ``=` `5``n ``=` `10` `print``(findNthTerm(a, b, c, n))` `# This code is contributed by mohit kumar 29`

## C#

 `// C# implementation of the approach``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 return the lcm of a and b``    ``static` `int` `lcm(``int` `a, ``int` `b)``    ``{``        ``return` `(a * b) / gcd(a, b);``    ``}` `    ``// Function to return the count of numbers``    ``// from 1 to num which are divisible by a, b or c``    ``static` `int` `divTermCount(``int` `a, ``int` `b, ``int` `c, ``int` `num)``    ``{` `        ``// Calculate number of terms divisible by a and``        ``// by b and by c then, remove the terms which is are``        ``// divisible by both a and b, both b and c, both``        ``// c and a and then add which are divisible by a and``        ``// b and c``        ``return` `((num / a) + (num / b) + (num / c)``                ``- (num / lcm(a, b))``                ``- (num / lcm(b, c))``                ``- (num / lcm(a, c))``                ``+ (num / lcm(a, lcm(b, c))));``    ``}` `    ``// Function to find the nth term``    ``// divisible by a, b or c``    ``// by using binary search``    ``static` `int` `findNthTerm(``int` `a, ``int` `b, ``int` `c, ``int` `n)``    ``{` `        ``// Set low to 1 and high to max (a, b, c) * n``        ``int` `low = 1, high = ``int``.MaxValue, mid;` `        ``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, c, 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 = 3, c = 5, n = 10;` `        ``Console.WriteLine(findNthTerm(a, b, c, n));` `    ``}``}` `/* This code is contributed by PrinciRaj1992 */`

## Javascript

 ``
Output:
`14`

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