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# Find Multiples of 2 or 3 or 5 less than or equal to N

• Last Updated : 23 Mar, 2021

Given an integer . The task is to count all such numbers that are less than or equal to N which are divisible by any of 2 or 3 or 5.
Note: If a number less than N is divisible by both 2 or 3, or 3 or 5, or all of 2,3 and 5 then also it should be counted only once.
Examples

Input : N = 5
Output : 4

Input : N = 10
Output : 8

Simple Approach: A simple approach is to traverse from 1 to N and count multiple of 2, 3, 5 which are less than equal to N. To do this, iterate up to N and just check whether a number is divisible by 2 or 3 or 5. If it is divisible, increment the counter and after reaching N, print the result.
Time Complexity: O(N).
Efficient Approach: An efficient approach is to use the concept of set theory. As we have to find numbers that are divisible by 2 or 3 or 5.

Now the task is to find n(a),n(b),n(c),n(ab), n(bc), n(ac), and n(abc). All these terms can be calculated using Bit masking. In this problem we have taken three numbers 2,3, and 5. So, the bit mask should be of 2^3 bits i.e 8 to generate all combination of 2,3, and 5.
Now according to the formula of set union, all terms containing odd numbers of (2,3,5) will add into the result and terms containing even number of (2,3,5) will get subtracted.
Below is the implementation of the above approach:

## C++

 // CPP program to count number of multiples// of 2 or 3 or 5 less than or equal to N #include  using namespace std; // Function to count number of multiples// of 2 or 3 or 5 less than or equal to Nint countMultiples(int n){    // As we have to check divisibility    // by three numbers, So we can implement    // bit masking    int multiple[] = { 2, 3, 5 };         int count = 0, mask = pow(2, 3);         for (int i = 1; i < mask; i++) {         // we check whether jth bit        // is set or not, if jth bit        // is set, simply multiply        // to prod        int prod = 1;                 for (int j = 0; j < 3; j++) {             // check for set bit            if (i & 1 << j)                prod = prod * multiple[j];        }                 // check multiple of product        if (__builtin_popcount(i) % 2 == 1)            count = count + n / prod;        else            count = count - n / prod;    }         return count;} // Driver codeint main(){    int n = 10;         cout << countMultiples(n) << endl;         return 0;}

## Java

 // Java program to count number of multiples// of 2 or 3 or 5 less than or equal to N class GFG{static int count_setbits(int N){    int cnt=0;    while(N>0)    {        cnt+=(N&1);        N=N>>1;    }    return cnt;} // Function to count number of multiples// of 2 or 3 or 5 less than or equal to Nstatic int countMultiples(int n){    // As we have to check divisibility    // by three numbers, So we can implement    // bit masking    int multiple[] = { 2, 3, 5 };         int count = 0, mask = (int)Math.pow(2, 3);         for (int i = 1; i < mask; i++) {         // we check whether jth bit        // is set or not, if jth bit        // is set, simply multiply        // to prod        int prod = 1;                 for (int j = 0; j < 3; j++) {             // check for set bit            if ((i & 1 << j)>0)                prod = prod * multiple[j];        }                 // check multiple of product        if (count_setbits(i) % 2 == 1)            count = count + n / prod;        else            count = count - n / prod;    }         return count;} // Driver codepublic static void main(String[] args){    int n = 10;         System.out.println(countMultiples(n));}}// this code is contributed by mits

## Python 3

 # Python3 program to count number of multiples# of 2 or 3 or 5 less than or equal to N  # Function to count number of multiples# of 2 or 3 or 5 less than or equal to Ndef countMultiples( n):     # As we have to check divisibility    # by three numbers, So we can implement    # bit masking    multiple = [ 2, 3, 5 ]         count = 0    mask = int(pow(2, 3))    for i in range(1,mask):        # we check whether jth bit        # is set or not, if jth bit        # is set, simply multiply        # to prod        prod = 1        for j in range(3):             # check for set bit            if (i & (1 << j)):                prod = prod * multiple[j]                 # check multiple of product        if (bin(i).count('1') % 2 == 1):            count = count + n // prod        else:            count = count - n // prod         return count  # Driver codeif __name__=='__main__':    n = 10    print(countMultiples(n))     # This code is contributed by ash264

## C#

 // C#  program to count number of multiples// of 2 or 3 or 5 less than or equal to N  using System; public class GFG{    static int count_setbits(int N){    int cnt=0;    while(N>0)    {        cnt+=(N&1);        N=N>>1;    }    return cnt;} // Function to count number of multiples// of 2 or 3 or 5 less than or equal to Nstatic int countMultiples(int n){    // As we have to check divisibility    // by three numbers, So we can implement    // bit masking    int []multiple = { 2, 3, 5 };         int count = 0, mask = (int)Math.Pow(2, 3);         for (int i = 1; i < mask; i++) {         // we check whether jth bit        // is set or not, if jth bit        // is set, simply multiply        // to prod        int prod = 1;                 for (int j = 0; j < 3; j++) {             // check for set bit            if ((i & 1 << j)>0)                prod = prod * multiple[j];        }                 // check multiple of product        if (count_setbits(i) % 2 == 1)            count = count + n / prod;        else            count = count - n / prod;    }         return count;} // Driver code    static public void Main (){             int n = 10;         Console.WriteLine(countMultiples(n));}}//This code is contributed by ajit.

## PHP

 > 1;    }     return $count;} // Function to count number of // multiples of 2 or 3 or 5 less// than or equal to Nfunction countMultiples($n){    // As we have to check divisibility    // by three numbers, So we can    // implement bit masking    $multiple = array(2, 3, 5);   $count = 0;    $mask = pow(2, 3);   for ($i = 1; $i < $mask; $i++) {  // we check whether jth bit // is set or not, if jth bit // is set, simply multiply // to prod $prod = 1;                 for ($j = 0; $j < 3; $j++) {  // check for set bit if ($i & 1 << $j) $prod = $prod * $multiple[$j]; }   // check multiple of product if (popcount($i) % 2 == 1)            $count = $count + (int)($n / $prod);                     else            $count = $count - (int)($n / $prod);                 }         return $count;} // Driver code$n = 10;     echo countMultiples(\$n);     // This code is contributed by ash264?>

## Javascript

 
Output:
8

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