Given a number X, find the Smallest Number, which is greater than 1, that divides XX. In the given question, X is always presumed to be greater than 1.
Examples:
Input: X = 6
Output: 2
Explanation: As, 66 is equal to 46656, which is divisible by 2 and it’s the smallest among all its divisors.Input: X = 3
Output: 3
Explanation: As, 33 is equal to 27, which is divisible by 3 and it’s the smallest among all its divisors.
Approach:
The main observation of this problem is that if a number P divides X, then it also divides XX, so we don’t need to calculate the value of XX. What we need to do is to find the smallest number that divides X which will always be a Prime Number.
Below is the implementation of the above approach:
C++
// C++ implementation of above approach #include <bits/stdc++.h> using namespace std; // Function to find the required smallest number int SmallestDiv(int n) { for (int i = 2; i * i <= n; i++) { // Finding smallest number that divides n if (n % i == 0) { // i divides n and return this // value immediately return i; } } // If n is a prime number then answer should be n, // As we can't take 1 as our answer. return n; } // Driver Code int main() { int X = 385; int ans = SmallestDiv(X); cout << ans << "\n"; return 0; } |
Java
// Java implementation of above approach class GFG{ // Function to find the // required smallest number static int SmallestDiv(int n) { for(int i = 2; i * i <= n; i++) { // Finding smallest number // that divides n if (n % i == 0) { // i divides n and return this // value immediately return i; } } // If n is a prime number then // answer should be n, as we // can't take 1 as our answer. return n; } // Driver Code public static void main(String[] args) { int X = 385; int ans = SmallestDiv(X); System.out.print(ans + "\n"); } } // This code is contributed by gauravrajput1 |
Python3
# Python3 implementation of above approach # Function to find the required smallest number def SmallestDiv(n): i = 2 while i * i <= n: # Finding smallest number that divides n if (n % i == 0): # i divides n and return this # value immediately return i i += 1 # If n is a prime number then answer should be n, # As we can't take 1 as our answer. return n # Driver Code if __name__=="__main__": X = 385 ans = SmallestDiv(X) print(ans) # This code is contributed by Yash_R |
C#
// C# implementation of above approach using System; class GFG { // Function to find the // required smallest number static int SmallestDiv(int n) { for(int i = 2; i * i <= n; i++) { // Finding smallest number // that divides n if (n % i == 0) { // i divides n and return this // value immediately return i; } } // If n is a prime number then // answer should be n, as we // can't take 1 as our answer. return n; } // Driver code public static void Main(String[] args) { int X = 385; int ans = SmallestDiv(X); Console.Write(ans + "\n"); } } // This code is contributed by shivanisinghss2110 |
Output: 5
Time Complexity: O(sqrt(X))
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