Given four integers **N**, **A**, **B**, and **C**. The task is to print the **N ^{th}** number in the set containing the multiples of

**A**,

**B**, or

**C**.

**Examples:**

Input:A = 2, B = 3, C = 5, N = 8Output:10

2, 3, 4, 5, 6, 8, 9,10, 12, 14, …Input:A = 2, B = 3, C = 5, N = 100Output:136

**Naive approach:** Start traversing from **1** until we find the **N ^{th}** 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

**N**number which is divisible by either

^{th}**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 <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 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)/gcd(a, b))* c) / gcd(((a*b)/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` |

## Javascript

`<script>` `// Javascript program to find nth term` `// divisible by a, b or c` `// Function to return` `// gcd of a and b` `function` `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` `function` `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` `parseInt(((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)/gcd(a, b))* c)/` ` ` `gcd(((a*b)/gcd(a, b)), c)))));` `}` `// Function for binary search to find the` `// nth term divisible by a, b or c` `function` `findNthTerm( a, b, c, n)` `{` ` ` `// Set low to 1 and high to LONG_MAX` ` ` `var` `low = 1, high = Number.MAX_SAFE_INTEGER , 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;` `}` `var` `a = 2, b = 3, c = 5, n = 100;` `document.write(parseInt(findNthTerm(a, b, c, n)));` `// This code is contributed by SoumikMondal` `</script>` |

**Output:**

136

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