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

• Last Updated : 19 Jul, 2022

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

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 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);``}` `long` `gcd_ab, gcd_bc, gcd_ac, lcm_abc, lcm_ab, lcm_bc, lcm_ac;` `void` `preCal(``long` `a, ``long` `b, ``long` `c) {``    ``gcd_ab = gcd(a, b); ``// GCD of a, b``    ``gcd_bc = gcd(c, b); ``// GCD of b, c``    ``gcd_ac = gcd(a, c); ``// GCD of a, c``    ``lcm_ab = ((a * b) / gcd_ab); ``// LCM of a, b``    ``lcm_bc = ((c * b) / gcd_bc); ``// LCM of b, c``    ``lcm_ac = ((a * c) / gcd_ac); ``// LCM of a, c``    ``lcm_abc = (lcm_ab * c) / gcd(lcm_ab, c);  ``// LCM of a, b, c``}` `// 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 / lcm_ab)``            ``- (num / lcm_bc)``            ``- (num / lcm_ac)``            ``+ (num / lcm_abc));``}` `// 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 (gcd(a, b, c) * n) and high to (max(a, b, c) * n)``      ``// (gcd(a, b, c) * n) is lowest possible value till index 'n'``      ``// (max(a, b, c) * n) is highest possible value till index 'n'``    ``preCal(a, b, c);``    ``long` `low = 1, high, mid;``    ``high = max(a, max(b, c)) * n;``    ``low = gcd(a, gcd_bc) * n;` `    ``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;``}` `// This code was improved by sharad_jain`

## Java

 `// Java program to find nth term``// divisible by a, b or c``class` `GFG``{``      ``static` `long` `gcd_ab, gcd_bc, gcd_ac, lcm_abc, lcm_ab, lcm_bc, lcm_ac;` `    ``// 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);``    ``}` `    ``static` `void` `preCal(``long` `a, ``long` `b, ``long` `c) {``        ``gcd_ab = gcd(a, b); ``// GCD of a, b``        ``gcd_bc = gcd(c, b); ``// GCD of b, c``        ``gcd_ac = gcd(a, c); ``// GCD of a, c``        ``lcm_ab = ((a * b) / gcd_ab); ``// LCM of a, b``        ``lcm_bc = ((c * b) / gcd_bc); ``// LCM of b, c``        ``lcm_ac = ((a * c) / gcd_ac); ``// LCM of a, c``        ``lcm_abc = (lcm_ab * c) / gcd(lcm_ab, c);  ``// LCM of a, b, c``    ``}` `    ``// 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 / lcm_ab)``                ``- (num / lcm_bc)``                ``- (num / lcm_ac)``                ``+ (num / lcm_abc));``    ``}` `    ``// 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 (gcd(a, b, c) * n) and high to (max(a, b, c) * n)``        ``// (gcd(a, b, c) * n) is lowest possible value till index 'n'``        ``// (max(a, b, c) * n) is highest possible value till index 'n'``        ``preCal(a, b, c);``        ``long` `low = ``1``, high, mid;``        ``high = a > b ? (a > c ? a : c) : (b > c ? b : c);``          ``high = high * n;``        ``low = gcd(a, gcd_bc) * n;` `        ``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``// This code was improved by sharad_jain`

## 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);``  ` `gcd_ab ``=` `1``; gcd_bc ``=` `1``; gcd_ac ``=` `1``; gcd_abc ``=` `1``;` `def` `preCal(a, b, c):``    ``gcd_ab ``=` `gcd(a, b);  ``# GCD of a, b``    ``gcd_bc ``=` `gcd(c, b);  ``# GCD of b, c``    ``gcd_ac ``=` `gcd(a, c);  ``# GCD of a, c``    ``gcd_abc ``=` `gcd(((a``*``b)``/``gcd_ab), c);  ``# GCD of a, b, c` `# 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_ab))``            ``-` `(num ``/` `((c ``*` `b) ``/` `gcd_bc))``            ``-` `(num ``/` `((a ``*` `c) ``/` `gcd_ac))``            ``+` `(num ``/` `((((a``*``b)``/``gcd_ab)``*` `c) ``/` `gcd_abc)));` `# Function for binary search to find the``# nth term divisible by a, b or c``def` `findNthTerm(a, b, c, n):` `    ``# Set low to (gcd(a, b, c) * n) and high to (max(a, b, c) * n)``      ``# (gcd(a, b, c) * n) is lowest possible value till index 'n'``      ``# (max(a, b, c) * n) is highest possible value till index 'n'``    ``preCal(a, b, c);``    ``mid ``=` `0``;``    ``high ``=` `max``(a, ``max``(b, c)) ``*` `n;``    ``low ``=` `gcd_abc ``*` `n;` `    ``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``# This code was improved by sharad_jain`

## C#

 `// C# program to find nth term``// divisible by a, b or c``using` `System;` `class` `GFG``{``      ``static` `long` `gcd_ab, gcd_bc, gcd_ac, lcm_abc, lcm_ab, lcm_bc, lcm_ac;` `    ``// 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);``    ``}` `    ``static` `void` `preCal(``long` `a, ``long` `b, ``long` `c) {``        ``gcd_ab = gcd(a, b); ``// GCD of a, b``        ``gcd_bc = gcd(c, b); ``// GCD of b, c``        ``gcd_ac = gcd(a, c); ``// GCD of a, c``        ``lcm_ab = ((a * b) / gcd_ab); ``// LCM of a, b``        ``lcm_bc = ((c * b) / gcd_bc); ``// LCM of b, c``        ``lcm_ac = ((a * c) / gcd_ac); ``// LCM of a, c``        ``lcm_abc = (lcm_ab * c) / gcd(lcm_ab, c);  ``// LCM of a, b, c``    ``}` `    ``// 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 / lcm_ab)``                ``- (num / lcm_bc)``                ``- (num / lcm_ac)``                ``+ (num / lcm_abc));``    ``}` `    ``// 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 (gcd(a, b, c) * n) and high to (max(a, b, c) * n)``        ``// (gcd(a, b, c) * n) is lowest possible value till index 'n'``        ``// (max(a, b, c) * n) is highest possible value till index 'n'``        ``preCal(a, b, c);``        ``long` `low = 1, high, mid;``        ``high = a > b ? (a > c ? a : c) : (b > c ? b : c);``          ``high = high * n;``        ``low = gcd(a, gcd_bc) * n;` `        ``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``// This code was improved by sharad_jain`

## Javascript

 ``

Output:

`136`

Time Compleixity: O(logn + log(min(a, b))

Auxiliary Space: O(1)

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