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

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++

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// 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 * 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

136


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Second year Department of Information Technology Jadavpur University

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Improved By : 29AjayKumar, princiraj1992