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

Given four integers **a**, **b**, **c** and **N**. The task is to find the **N ^{th}** 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 **N ^{th}** 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 <bits/stdc++.h> ` `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; ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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 */` |

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**Output:**

14

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