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In this problem, we are given a number N. Our task is to Find Multiples of 2 or 3 or 5 less than or equal to N.

**Problem Description** − We will be counting all elements from 1 to N that are divisible by 2 or 3 or 5.

**Let’s take an example to understand the problem,**

N = 7

5

All the elements from 1 to 7 are : 1, 2, 3, 4, 5, 6, 7. Elements divisible by 2/3/5 are 2, 3, 4, 5, 6

A simple approach to solve the problem is by traversing all numbers from 1 to N and count all numbers that are divided by 2 or 3 or 5.

**Initialize** − count = 0

**Step 1** − loop for i = 1 to N.

**Step 1.1**: if(i%2 == 0 || i%3 == 0 || i%5 == 0), count++.

**Step 2** − return count.

A more effective approach to solve the problem is using the set theory.

The count of number divisible by 2 is n(2)

The count of number divisible by 3 is n(3)

The count of number divisible by 5 is n(5)

The count of number divisible by 2 and 3 is n(2 n 3)

The count of number divisible by 2 and 5 is n(2 n 5)

The count of number divisible by 3 and 5 is n(3 n 5)

The count of number divisible by 2 and 3 and 5 is n(2 n 3 n 5)

The count of number divisible by 2 or 3 or 5 is n(2 U 3 U 5)

Based on set theory,

n(2 ∪ 3 ∪ 5) = n(2) + n(3) + n(5) - n(2 ∩ 3) - n(2 ∩ 5) - n(3 ∩ 5) + n(2 ∩ 3 ∩ 5)

The solution is found by calculating the bit masks of numbers.

**Program to illustrate the working of our solution,**

#include <bits/stdc++.h> using namespace std; int countMultiples(int n) { int values[] = { 2, 3, 5 }; int countMultiples = 0, bitMask = pow(2, 3); for (int i = 1; i < bitMask; i++) { int prod = 1; for (int j = 0; j < 3; j++) { if (i & 1 << j) prod = prod * values[j]; } if (__builtin_popcount(i) % 2 == 1) countMultiples = countMultiples + n / prod; else countMultiples = countMultiples - n / prod; } return countMultiples; } int main() { int n = 13; cout<<"The number of multiples till "<<n<<" is "<<countMultiples(n)<<endl; return 0; }

The number of multiples till 13 is 9

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