Count of Multiples of A ,B or C less than or equal to N

Given four integers N, A, B and C. The task is to find the count of integers from the range [1, N] which are divisible by either A, B or C.

Examples:

Input: A = 2, B = 3, C = 5, N = 10
Output: 8
2, 3, 4, 5, 6, 8, 9 and 10 are the only number from the
range [1, 10] which are divisible by wither 2, 3 or 5.



Input: A = 7, B = 3, C = 5, N = 100
Output: 55

Approach: An efficient approach is to use the concept of set theory. As we have to find numbers that are divisible by a or b or c.
  \begin{document} \begin{itemize} \item Let $n(a) \colon $ count of numbers divisible by a. \item Let $n(b) \colon $ count of numbers divisible by b. \item Let $n(c) \colon $ count of numbers divisible by c. \item $n(a \bigcap b) \colon $ count of numbers divisible by a and b. \item $n(a \bigcap c) \colon $ count of numbers divisible by b and c. \item $n(b \bigcap c) \colon $ count of numbers divisible by c and a. \item $n(a \bigcap b \bigcap c) \colon $ count of numbers divisible by a and b and c. \end{itemize}  According to set theory,  $n\left( a \bigcup b \bigcup c \right)=n(a)+n(b)+n(c)-n(a \bigcap b)-n(b \bigcap c)-n(a \bigcap c)+n(a \bigcap b \bigcap c)$ \end{document}
So. the count of numbers divisible either by A, B or C is (num/A) + (num/B) + (num/C) – (num/lcm(A, B)) – (num/lcm(A, B)) – (num/lcm(A, C)) + – (num/lcm(A, B, C))

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the
// gcd of a and b
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
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))));
}
  
// Driver code
int main()
{
    long a = 7, b = 3, c = 5, n = 100;
  
    cout << divTermCount(a, b, c, n);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
import java.util.*;
      
class GFG
{
      
// Function to return the
// 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))));
}
  
// Driver code
static public void main (String []arr)
{
    long a = 7, b = 3, c = 5, n = 100;
  
    System.out.println(divTermCount(a, b, c, n));
}
}
  
// This code is contributed by 29AjayKumar

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach 
  
# Function to return the 
# 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)))); 
  
# Driver code 
if __name__ == "__main__"
  
    a = 7; b = 3; c = 5; n = 100
  
    print(divTermCount(a, b, c, n)); 
  
# This code is contributed by AnkitRai01

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation for above approach
using System;
      
class GFG
{
      
// Function to return the
// 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))));
}
  
// Driver code
static public void Main (String []arr)
{
    long a = 7, b = 3, c = 5, n = 100;
  
    Console.WriteLine(divTermCount(a, b, c, n));
}
}
  
// This code is contributed by 29AjayKumar

chevron_right


Output:

55


My Personal Notes arrow_drop_up

Second year Department of Information Technology Jadavpur University

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : AnkitRai01, 29AjayKumar