Given an integer **N**. The task is to find a number that is smaller than or equal to N and has maximum prime factors. In case there are two or more numbers with same maximum number of prime factors, find the smallest of all.

Examples:

Input : N = 10 Output : 6 Number of prime factor of: 1 : 0 2 : 1 3 : 1 4 : 1 5 : 1 6 : 2 7 : 1 8 : 1 9 : 1 10 : 2 6 and 10 have maximum (2) prime factor but 6 is smaller. Input : N = 40 Output : 30

**Method 1 (brute force):**

For each integer from 1 to N, find the number of prime factor of each integer and find the smallest number having maximum number of prime factors.

**Method 2 (Better Approach):**

Use sieve method to count number of prime factor of each number less than N. And find the minimum number having maximum count.

Below is C++ implementation of this approach:

// C++ program to find integer having maximum number // of prime factor in first N natural numbers. #include<bits/stdc++.h> using namespace std; // Return smallest number having maximum // prime factors. int maxPrimefactorNum(int N) { int arr[N + 5]; memset(arr, 0, sizeof(arr)); // Sieve of eratosthenes method to count // number of prime factors. for (int i = 2; i*i <= N; i++) { if (!arr[i]) for (int j = 2*i; j <= N; j+=i) arr[j]++; arr[i] = 1; } int maxval = 0, maxint = 1; // Finding number having maximum number // of prime factor. for (int i = 1; i <= N; i++) { if (arr[i] > maxval) { maxval = arr[i]; maxint = i; } } return maxint; } // Driven Program int main() { int N = 40; cout << maxPrimefactorNum(N) << endl; return 0; }

Output:

30

**Method 3 (efficient approach):**

Generate all prime number before N using Sieve. Now, multiply consecutive prime numbers (starting from first prime number) one after another until the product is less than N.

Below is C++ implementation of this approach:

// C++ program to find integer having maximum number // of prime factor in first N natural numbers #include<bits/stdc++.h> using namespace std; // Return smallest number having maximum prime factors. int maxPrimefactorNum(int N) { bool arr[N + 5]; memset(arr, true, sizeof(arr)); // Sieve of eratosthenes for (int i = 3; i*i <= N; i += 2) { if (!arr[i]) for (int j = i*i; j <= N; j+=i) arr[j] = false; } // Storing prime numbers. vector<int> prime; prime.push_back(2); for(int i = 3; i <= N; i += 2) if(arr[i]) prime.push_back(i); // Generating number having maximum prime factors. int i = 0, ans = 1; while (ans*prime[i] <= N && i < prime.size()) { ans *= prime[i]; i++; } return ans; } // Driven Program int main() { int N = 40; cout << maxPrimefactorNum(N) << endl; return 0; }

Output:

30

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