# Nth number in a set of multiples of A , B or C

Given four integers N, A, B and C. The task is to print the Nth number in the set containing the multiples of A, B or C.

Examples:

Input: A = 2, B = 3, C = 5, N = 8
Output: 10
2, 3, 4, 5, 6, 8, 9, 10, 12, 14, …

Input: A = 2, B = 3, C = 5, N = 100
Output: 136

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

Naive approach: Start traversing from 1 until we find the Nth element which is divisible by either A, B or C.

Efficient approach: Given a number we can find the count of the divisors of either A, B or C. Now, binary search can be used to find the Nth number which is divisible by either A, B or C.

So, if the number is num then
count = (num/A) + (num/B) + (num/C) – (num/lcm(A, B)) – (num/lcm(C, B)) – (num/lcm(A, C)) – (num/lcm(A, B, C))

Below is the implementation of the above approach:

## C++

 `// C++ program to find nth term ` `// divisible by a, b or c ` ` `  `#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 count of integers ` `// from the range [1, num] which are ` `// divisible by either a, b or c ` `long` `divTermCount(``long` `a, ``long` `b, ``long` `c, ``long` `num) ` `{ ` `    ``// Calculate the number of terms divisible by a, b ` `    ``// and c then remove the terms which are divisible ` `    ``// by both (a, b) or (b, c) or (c, a) and then ` `    ``// add the numbers which are divisible by a, b and c ` `    ``return` `((num / a) + (num / b) + (num / c) ` `            ``- (num / ((a * b) / gcd(a, b))) ` `            ``- (num / ((c * b) / gcd(c, b))) ` `            ``- (num / ((a * c) / gcd(a, c))) ` `            ``+ (num / ((a * b * c) / gcd(gcd(a, b), c)))); ` `} ` ` `  `// Function for binary search to find the ` `// nth term divisible by a, b or c ` `int` `findNthTerm(``int` `a, ``int` `b, ``int` `c, ``long` `n) ` `{ ` `    ``// Set low to 1 and high to LONG_MAX ` `    ``long` `low = 1, high = LONG_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() ` `{ ` `    ``long` `a = 2, b = 3, c = 5, n = 100; ` ` `  `    ``cout << findNthTerm(a, b, c, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find nth term ` `// divisible by a, b or c ` `class` `GFG ` `{ ` ` `  `    ``// Function to return ` `    ``// gcd of a and b ` `    ``static` `long` `gcd(``long` `a, ``long` `b) ` `    ``{ ` `        ``if` `(a == ``0``) ` `        ``{ ` `            ``return` `b; ` `        ``} ` `        ``return` `gcd(b % a, a); ` `    ``} ` ` `  `    ``// Function to return the count of integers ` `    ``// from the range [1, num] which are ` `    ``// divisible by either a, b or c ` `    ``static` `long` `divTermCount(``long` `a, ``long` `b,  ` `                             ``long` `c, ``long` `num)  ` `    ``{ ` `        ``// Calculate the number of terms divisible by a, b ` `        ``// and c then remove the terms which are divisible ` `        ``// by both (a, b) or (b, c) or (c, a) and then ` `        ``// add the numbers which are divisible by a, b and c ` `        ``return` `((num / a) + (num / b) + (num / c) -  ` `                ``(num / ((a * b) / gcd(a, b))) -  ` `                ``(num / ((c * b) / gcd(c, b))) -  ` `                ``(num / ((a * c) / gcd(a, c))) +  ` `                ``(num / ((a * b * c) / gcd(gcd(a, b), c)))); ` `    ``} ` ` `  `    ``// Function for binary search to find the ` `    ``// nth term divisible by a, b or c ` `    ``static` `long` `findNthTerm(``int` `a, ``int` `b, ``int` `c, ``long` `n)  ` `    ``{ ` `         `  `        ``// Set low to 1 and high to LONG_MAX ` `        ``long` `low = ``1``, high = Long.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 = ``100``; ` ` `  `        ``System.out.println(findNthTerm(a, b, c, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python3 program to find nth term ` `# divisible by a, b or c ` `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 return the count of integers ` `# from the range [1, num] which are ` `# divisible by either a, b or c ` `def` `divTermCount(a, b, c, num): ` `     `  `    ``# Calculate the number of terms divisible by a, b ` `    ``# and c then remove the terms which are divisible ` `    ``# by both (a, b) or (b, c) or (c, a) and then ` `    ``# add the numbers which are divisible by a, b and c ` `    ``return` `((num ``/` `a) ``+` `(num ``/` `b) ``+` `(num ``/` `c) ``-`  `                ``(num ``/` `((a ``*` `b) ``/` `gcd(a, b))) ``-`  `                ``(num ``/` `((c ``*` `b) ``/` `gcd(c, b))) ``-`  `                ``(num ``/` `((a ``*` `c) ``/` `gcd(a, c))) ``+`  `                ``(num ``/` `((a ``*` `b ``*` `c) ``/` `gcd(gcd(a, b), c)))); ` ` `  `# Function for binary search to find the ` `# nth term divisible by a, b or c ` `def` `findNthTerm(a, b, c, n): ` ` `  `    ``# Set low to 1 and high to LONG_MAX ` `    ``low ``=` `1``; high ``=` `sys.maxsize; 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` `int``(low); ` ` `  `# Driver code ` `a ``=` `2``; b ``=` `3``; c ``=` `5``; n ``=` `100``; ` ` `  `print``(findNthTerm(a, b, c, n)); ` ` `  `# This code is contributed by 29AjayKumar `

## C#

 `// C# program to find nth term ` `// divisible by a, b or c ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `    ``// Function to return ` `    ``// gcd of a and b ` `    ``static` `long` `gcd(``long` `a, ``long` `b) ` `    ``{ ` `        ``if` `(a == 0) ` `        ``{ ` `            ``return` `b; ` `        ``} ` `        ``return` `gcd(b % a, a); ` `    ``} ` ` `  `    ``// Function to return the count of integers ` `    ``// from the range [1, num] which are ` `    ``// divisible by either a, b or c ` `    ``static` `long` `divTermCount(``long` `a, ``long` `b,  ` `                             ``long` `c, ``long` `num)  ` `    ``{ ` `        ``// Calculate the number of terms divisible by a, b ` `        ``// and c then remove the terms which are divisible ` `        ``// by both (a, b) or (b, c) or (c, a) and then ` `        ``// add the numbers which are divisible by a, b and c ` `        ``return` `((num / a) + (num / b) + (num / c) -  ` `                ``(num / ((a * b) / gcd(a, b))) -  ` `                ``(num / ((c * b) / gcd(c, b))) -  ` `                ``(num / ((a * c) / gcd(a, c))) +  ` `                ``(num / ((a * b * c) / gcd(gcd(a, b), c)))); ` `    ``} ` ` `  `    ``// Function for binary search to find the ` `    ``// nth term divisible by a, b or c ` `    ``static` `long` `findNthTerm(``int` `a, ``int` `b,  ` `                            ``int` `c, ``long` `n)  ` `    ``{ ` `         `  `        ``// Set low to 1 and high to LONG_MAX ` `        ``long` `low = 1, high = ``long``.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 = 100; ` ` `  `        ``Console.WriteLine(findNthTerm(a, b, c, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:

```136
```

My Personal Notes arrow_drop_up Second year Department of Information Technology Jadavpur University

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Improved By : 29AjayKumar, princiraj1992