Given a number ‘n’, check whether it is an emirpimes or not.
An emirpimes(“semiprime” when spelled backwards) derives its definition from the way it is spelt. So, an emirpimes is a number that is a semiprime (product of two prime numbers) itself, and the reversal of its digits gives another new number, which too is a semi prime. Hence, by definition we can conclude that none of the palindrome numbers can be emirpimes, as the reversal of their digits does not give any new number, but forms the same number again.
Examples :
Input : 15
Output : Yes
Explanation : 15 is itself a semi prime number, since it is a product of two prime numbers 3 and 5. The reversal of its digits gives a new number 51, which too is a semi prime, it being the product of two prime numbers, viz., 3 and 17Input : 49
Output : Yes
Explanation : 49 is itself a semi prime number, since it is a product of two prime numbers(not necessarily distinct) 7 and 7. The reversal of its digits gives a new number 94, which too is a semi prime, it being the product of two prime numbers, viz., 2 and 47Input : 25
Output : No
Explanation : 25 is itself a semi prime number, since it is a product of two prime numbers(not necessarily distinct) 5 and 5. The reversal of its digits gives a new number 52, which is not a semi prime, it being the product of three and not two prime numbers, viz., 2, 2 and 13
Approach :
- First check whether the entered number is semi prime, itself.
- If yes, form a number by reversing its digits.
- Now, compare this number with the initially entered number to ascertain if the number is palindrome or not.
- If the number is not a palindrome, check whether this new number is also semi prime or not.
- If yes, then the initially entered number is reported to be an emirpimes.
C++
// CPP code to check whether// a number is Emirpimes or not#include <bits/stdc++.h>using namespace std;// Checking whether a number// is semi-prime or notint checkSemiprime(int num){ int cnt = 0; for (int i = 2; cnt < 2 && i * i <= num; ++i) { while (num % i == 0) { num /= i; // Increment count of // prime numbers ++cnt; } } // If number is still greater than 1, after // exiting the for loop add it to the count // variable as it indicates the number is // a prime number if (num > 1) ++cnt; // Return '1' if count is // equal to '2' else return '0' return cnt == 2;}// Checking whether a number// is emirpimes or notbool isEmirpimes(int n){ // Number itself is not semiprime. if (checkSemiprime(n) == false) return false; // Finding reverse of n. int r = 0; for (int t=n; t!=0; t=t/n) r = r * 10 + t % 10; // The definition of emirpimes excludes // palindromes, hence we do not check // further, if the number entered is a // palindrome if (r == n) return false; // Checking whether the reverse of the // semi prime number entered is also // a semi prime number or not return (checkSemiprime(r));}// Driver Codeint main(){ int n = 15; if (isEmirpimes(n)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java code to check whether a// number is Emirpimes or notimport java.io.*;class GFG { // Checking whether a number // is semi-prime or not static boolean checkSemiprime(int num) { int cnt = 0; for (int i = 2; cnt < 2 && i * i <= num; ++i) { while (num % i == 0) { num /= i; // Increment count of // prime numbers ++cnt; } } // If number is still greater than 1, // after exiting the for loop add it // to the count variable as it indicates // the number is a prime number if (num > 1) ++cnt; // Return '1' if count is equal // to '2' else return '0' return cnt == 2; } // Checking whether a number // is emirpimes or not static boolean isEmirpimes(int n) { // Number itself is not semiprime. if (checkSemiprime(n) == false) return false; // Finding reverse of n. int r = 0; for (int t = n; t != 0; t = t / n) r = r * 10 + t % 10; // The definition of emirpimes excludes // palindromes, hence we do not check // further, if the number entered is a // palindrome if (r == n) return false; // Checking whether the reverse of the // semi prime number entered is also // a semi prime number or not return (checkSemiprime(r)); } // Driver Code public static void main (String[] args) { int n = 15; if (isEmirpimes(n)) System.out.println("Yes"); else System.out.println("No"); }}// This code is contributed by Ajit. |
Python3
# Python3 code to check whether # a number is Emirpimesor not # Checking whether a number # is semi-prime or not def checkSemiprime(num): cnt = 0; i = 2; while (cnt < 2 and (i * i) <= num): while (num % i == 0): num /= i; # Increment count of # prime numbers cnt += 1; i += 1; # If number is still greater than 1, # after exiting the add it to the # count variable as it indicates # the number is a prime number if (num > 1): cnt += 1; # Return '1' if count is equal # to '2' else return '0' return cnt == 2; # Checking whether a number # is emirpimes or not def isEmirpimes(n): # Number itself is not semiprime. if (checkSemiprime(n) == False): return False; # Finding reverse of n. r = 0; t = n; while (t != 0): r = r * 10 + t % 10; t = t / n; # The definition of emirpimes excludes # palindromes, hence we do not check # further, if the number entered # is a palindrome if (r == n): return false; # Checking whether the reverse of the # semi prime number entered is also # a semi prime number or not return (checkSemiprime(r)); # Driver Code n = 15; if (isEmirpimes(n)): print("No"); else: print("Yes"); # This code is contributed by mits |
C#
// C# code to check whether a// number is Emirpimes or notusing System;class GFG{ // Checking whether a number // is semi-prime or not static bool checkSemiprime(int num) { int cnt = 0; for (int i = 2; cnt < 2 && i * i <= num; ++i) { while (num % i == 0) { num /= i; // Increment count of // prime numbers ++cnt; } } // If number is still greater than 1, // after exiting the for loop add it // to the count variable as it // indicates the number is a prime number if (num > 1) ++cnt; // Return '1' if count is equal // to '2' else return '0' return cnt == 2; } // Checking whether a number // is emirpimes or not static bool isEmirpimes(int n) { // Number itself is not semiprime. if (checkSemiprime(n) == false) return false; // Finding reverse of n. int r = 0; for (int t = n; t != 0; t = t / n) r = r * 10 + t % 10; // The definition of emirpimes excludes // palindromes, hence we do not check // further, if the number entered is a // palindrome if (r == n) return false; // Checking whether the reverse of the // semi prime number entered is also // a semi prime number or not return (checkSemiprime(r)); } // Driver Code public static void Main () { int n = 15; if (isEmirpimes(n)) Console.WriteLine("Yes"); else Console.WriteLine("No"); }}// This code is contributed by anuj_67. |
PHP
<?php// PHP code to check whether // a number is Emirpimesor not// Checking whether a number// is semi-prime or notfunction checkSemiprime($num){ $cnt = 0; for ($i = 2; $cnt < 2 && $i * $i <= $num; ++$i) { while ($num % $i == 0) { $num /= $i; // Increment count of // prime numbers ++$cnt; } } // If number is still greater // than 1, after exiting the // for loop add it to the // count variable as it // indicates the number is a // prime number if ($num > 1) ++$cnt; // Return '1' if count // is equal to '2' else // return '0' return $cnt == 2;}// Checking whether a number// is emirpimes or notfunction isEmirpimes($n){ // Number itself is // not semiprime. if (checkSemiprime($n) == false) return false; // Finding reverse // of n. $r = 0; for ($t = $n; $t != 0; $t = $t / $n) $r = $r * 10 + $t % 10; // The definition of emirpimes // excludes palindromes,hence // we do not check further, // if the number entered // is a palindrome if ($r == $n) return false; // Checking whether the // reverse of the // semi prime number // entered is also // a semi prime number // or not return (checkSemiprime($r));} // Driver Code $n = 15; if (isEmirpimes($n)) echo "No"; else echo "Yes";// This code is contributed by Ajit.?> |
Javascript
<script>// Javascript code to check whether a// number is Emirpimes or not// Checking whether a number// is semi-prime or notfunction checkSemiprime(num){ let cnt = 0; for(let i = 2; cnt < 2 && i * i <= num; ++i) { while (num % i == 0) { num = parseInt(num / i, 10); // Increment count of // prime numbers ++cnt; } } // If number is still greater than 1, // after exiting the for loop add it // to the count variable as it // indicates the number is a prime number if (num > 1) ++cnt; // Return '1' if count is equal // to '2' else return '0' return cnt == 2;} // Checking whether a number// is emirpimes or notfunction isEmirpimes(n){ // Number itself is not semiprime. if (checkSemiprime(n) == false) return false; // Finding reverse of n. let r = 0; for(let t = n; t != 0; t = parseInt(t / n, 10)) r = r * 10 + t % 10; // The definition of emirpimes excludes // palindromes, hence we do not check // further, if the number entered is a // palindrome if (r == n) return false; // Checking whether the reverse of the // semi prime number entered is also // a semi prime number or not return (checkSemiprime(r));}// Driver codelet n = 15;if (isEmirpimes(n)) document.write("Yes");else document.write("No"); // This code is contributed by decode2207</script> |
Output :
Yes
Share your thoughts in the comments
