Magnanimous Number is a number of at least 2 digits such that the sum obtained inserting a “+” among its digit in any position gives a prime.
For example:
4001 is Magnanimous Number because the numbers 4+001=5, 40+01=41 and 400+1=401 are all prime numbers.
Check if N is a Magnanimous number
Given a number N, the task is to check if N is an Magnanimous Number or not. If N is a Magnanimous Number then print “Yes” else print “No”.
Examples:
Input: N = 4001
Output: Yes
Explanation:
4+001=5, 40+01=41 and 400+1=401 are all prime numbers.Input: N = 18
Output: No
Approach:
- Convert the number N to string
- Traverse the string and find all left part and right part of the string.
- Convert the left part and right part of the string to integer and check if the sum of left part and right part is not a prime number then return false
- Otherwise, return true at last
For example if N = 4001
left part + right part = prime number
4+001=5 = prime number
40+01=41 prime number
400+1=401 prime number
Below is the implementation of the above approach:
C++
// C++ implementation to check // if a number is Magnanimous#include <bits/stdc++.h>using namespace std;// Function to check if n is primebool isPrime(int n){ // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false; for (int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true;}// Function to check if the number is // Magnanimous or notbool isMagnanimous(int N){ // converting the number to string string s = to_string(N); // finding length of string int l = s.length(); // number should not be of single digit if (l < 2) return false; // loop to find all left and right // part of the string for (int i = 0; i < l - 1; i++) { string left = s.substr(0, i + 1); string right = s.substr(i + 1); int x = stoi(left); int y = stoi(right); if (!isPrime(x + y)) return false; } return true;}// Driver Codeint main(){ int N = 12; isMagnanimous(N) ? cout << "Yes" : cout << "No"; return 0;} |
Java
// Java implementation to check // if a number is Magnanimous class GFG{ // Function to check if n is primestatic boolean isPrime(int n){ // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false; for(int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true;}// Function to check if the number is // Magnanimous or notstatic boolean isMagnanimous(int N){ // Converting the number to string String s = Integer.toString(N); // Finding length of string int l = s.length(); // Number should not be of single digit if (l < 2) return false; // Loop to find all left and right // part of the string for(int i = 0; i < l - 1; i++) { String left = s.substring(0, i + 1); String right = s.substring(i + 1); int x = Integer. valueOf(left); int y = Integer. valueOf(right); if (!isPrime(x + y)) return false; } return true;}// Driver code public static void main(String[] args) { int N = 12; if(isMagnanimous(N)) System.out.print("Yes\n"); else System.out.print("No\n"); } } // This code is contributed by shubham |
Python3
# Python3 implementation to check # if a number is Magnanimous # Function to check if n is prime def isPrime(n): # Corner cases if (n <= 1): return False if (n <= 3): return True # This is checked so that we can skip # middle five numbers in below loop if (n % 2 == 0) or (n % 3 == 0): return False i = 5 while (i * i <= n): if (n % i == 0 or n % (i + 2) == 0): return False i = i + 6 return True# Function to check if the number is # Magnanimous or not def isMagnanimous(N): # Converting the number to string s = str(N) # Finding length of string l = len(s) # Number should not be of single digit if (l < 2): return False # Loop to find all left and right # part of the string for i in range(l - 1): left = s[0 : i + 1] right = s[i + 1 : ] x = int(left) y = int(right) if (not isPrime(x + y)): return False return True# Driver codeN = 12if isMagnanimous(N): print("Yes")else: print("No")# This code is contributed by divyeshrabadiya07 |
C#
// C# implementation to check // if a number is Magnanimous using System;class GFG{ // Function to check if n is primestatic bool isPrime(int n){ // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false; for(int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true;}// Function to check if the number is // Magnanimous or notstatic bool isMagnanimous(int N){ // Converting the number to string String s = N.ToString(); // Finding length of string int l = s.Length; // Number should not be of single digit if (l < 2) return false; // Loop to find all left and right // part of the string for(int i = 0; i < l - 1; i++) { String left = s.Substring(0, i + 1); String right = s.Substring(i + 1); int x = int.Parse(left); int y = int. Parse(right); if (!isPrime(x + y)) return false; } return true;}// Driver code public static void Main(String[] args) { int N = 12; if(isMagnanimous(N)) Console.Write("Yes\n"); else Console.Write("No\n"); } } // This code is contributed by amal kumar choubey |
Javascript
<script> // Javascript implementation to check // if a number is Magnanimous // Function to check if n is prime function isPrime(n) { // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false; for(let i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true; } // Function to check if the number is // Magnanimous or not function isMagnanimous(N) { // Converting the number to string let s = N.toString(); // Finding length of string let l = s.length; // Number should not be of single digit if (l < 2) return false; // Loop to find all left and right // part of the string for(let i = 0; i < l - 1; i++) { let left = s.substring(0, i + 1); let right = s.substring(i + 1); let x = parseInt(left); let y = parseInt(right); if (!isPrime(x + y)) return false; } return true; } let N = 12; if(isMagnanimous(N)) document.write("Yes"); else document.write("No");// This code is contributed by mukesh07.</script> |
Output:
Yes
Time Complexity: O(n)
Reference: http://www.numbersaplenty.com/set/magnanimous_number/
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