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

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  • 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 <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);
}
 
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




<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);
}
 
var gcd_ab, gcd_bc, gcd_ac, lcm_abc, lcm_ab, lcm_bc, lcm_ac;
 
function 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
    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
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 / 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
function 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);
    var 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;
}
 
var a = 2, b = 3, c = 5, n = 100;
document.write(parseInt(findNthTerm(a, b, c, n)));
 
 
// This code is contributed by SoumikMondal
// This code was improved by sharad_jain
 
</script>

Output: 

136

 

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

Auxiliary Space: O(1)


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