Program to find Nth term divisible by a or b in C++

In this problem, we are given three numbers A, B, and N. Our task is to create a program to find Nth term divisible by A or B in C++.

Problem Description

The Nth Term divisible by A or B. Here, we will find the term n number term which is divisible by number A or B. For this, we will count till nth numbers that are divisible by A or B.

Let’s take an example to understand the problem,

Input

A = 4, B = 3, N = 5

Output

9

Explanation

The terms the are divisible by 3 and 4 are −

3, 4, 6, 8, 9, 12, …

5th term is 9.

Solution Approach

To find the nth term that is divisible by A or B. We can simply find numbers divisible by A or B and the nth term in the series is our answer.

Example

 Live Demo

#include<iostream>
using namespace std;
int findNTerm(int N, int A, int B) {
   int count = 0;
   int num = 1;
   while( count < N){
      if(num%A == 0 || num%B == 0)
         count++;
      if(count == N)
         return num;
         num++;
   }
   return 0;
}
int main(){
   int N = 12, A = 3, B = 4;
   cout<<N<<"th term divisible by "<<A<<" or "<<B<<" is "<<findNTerm(N, A, B)<<endl;
}

Output

12th term divisible by 3 or 4 is 24

Another approach to solving the problem could be using the Binary search to find the Nth element which is divisible by A or B. We will find the Nth term using the formula−

NTerm = maxNum/A + maxNum/B + maxNum/lcm(A,B)

And based on the value of Nterm, we will check if we need to traverse in the numbers less than maxNum or greater than maxNum.

Example

 Live Demo

#include <iostream>
using namespace std;
int findLCM(int a, int b) {
   int LCM = a, i = 2;
   while(LCM % b != 0) {
      LCM = a*i;
      i++;
   }
   return LCM;
}
int findNTerm(int N, int A, int B) {
   int start = 1, end = (N*A*B), mid;
   int LCM = findLCM(A, B);

while (start < end) {
   mid = start + (end - start) / 2;
   if ( ((mid/A) + (mid/B) - (mid/LCM)) < N)
      start = mid + 1;
   else
      end = mid;
}
   return start;
}
int main() {
   int N = 12, A = 3, B = 4;
   cout<<N<<"th term divisible by "<<A<<" or "<<B<<" is "<<findNTerm(N, A, B);
}

Output

12th term divisible by 3 or 4 is 24