Given an integer N, the task is to check if N is a Rhonda numbers to base 10.
Rhonda numbers to base 10 are numbers if the product of its digits equals 10*Sum of prime factors of N (including multiplicity).
Examples:
Input: N = 1568
Output: Yes
Explanation:
1568’s prime factorization = 25 * 72.
Sum of prime factors = 2*5+7*2=24.
Product of digits of 1568 = 1*5*6*8=240 = 10*24.
Hence 1568 is a Rhonda number to base 10.Input: N = 28
Output: No
Approach: The idea is to find the sum of all prime factors of N and check if ten times of Sum of prime factors of N is equal to the product of digits of N or not.
Below is the implementation of the above approach:
C++
// C++ implementation to check if N// is a Rhonda number#include <bits/stdc++.h>using namespace std;// Function to find the // product of digitsint getProduct(int n){ int product = 1; while (n != 0) { product = product * (n % 10); n = n / 10; } return product;}// Function to find sum of all prime// factors of a given number Nint sumOfprimeFactors(int n){ int sum = 0; // add the number of // 2s that divide n while (n % 2 == 0) { sum += 2; n = n / 2; } // N must be odd at this // point. So we can skip // one element for (int i = 3; i <= sqrt(n); i = i + 2) { // While i divides n, // add i and divide n while (n % i == 0) { sum += i; n = n / i; } } // Condition to handle the case when N // is a prime number greater than 2 if (n > 2) sum += n; return sum;}// Function to check if n// is Rhonda numberbool isRhondaNum(int N){ return 10 * sumOfprimeFactors(N) == getProduct(N);}// Driver codeint main(){ int n = 1568; if (isRhondaNum(n)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java program to check if N // is a Rhonda number import java.util.*;import java.lang.*; // import Math;class GFG{// Function to find the // product of digits static int getProduct(int n) { int product = 1; while (n != 0) { product = product * (n % 10); n = n / 10; } return product; } // Function to find sum of all prime // factors of a given number N static int sumOfprimeFactors(int n) { int sum = 0; // Add the number of // 2s that divide n while (n % 2 == 0) { sum += 2; n = n / 2; } // N must be odd at this // point. So we can skip // one element for(int i = 3; i <= Math.sqrt(n); i = i + 2) { // While i divides n, // add i and divide n while (n % i == 0) { sum += i; n = n / i; } } // Condition to handle the case when N // is a prime number greater than 2 if (n > 2) sum += n; return sum; }// Function to check if n // is Rhonda number static boolean isRhondaNum(int N) { return (10 * sumOfprimeFactors(N) == getProduct(N));}// Driver Codepublic static void main(String[] args){ // Given Number n int n = 1568; if (isRhondaNum(n)) { System.out.println("Yes"); } else { System.out.println("No"); }}}// This code is contributed by vikas_g |
Python3
# Python3 implementation to check# if N is a Rhonda number import math# Function to find the # product of digits def getProduct(n): product = 1 while (n != 0): product = product * (n % 10) n = n // 10 return product# Function to find sum of all prime # factors of a given number N def sumOfprimeFactors(n): Sum = 0 # Add the number of # 2s that divide n while (n % 2 == 0): Sum += 2 n = n // 2 # N must be odd at this # point. So we can skip # one element for i in range(3, int(math.sqrt(n)) + 1, 2): # While i divides n, # add i and divide n while (n % i == 0): Sum += i n = n // i # Condition to handle the case when N # is a prime number greater than 2 if (n > 2): Sum += n return Sum# Function to check if n # is Rhonda number def isRhondaNum(N): return bool(10 * sumOfprimeFactors(N) == getProduct(N)) # Driver coden = 1568if (isRhondaNum(n)): print("Yes") else: print("No")# This code is contributed by divyeshrabadiya07 |
C#
// C# implementation to check if N // is a Rhonda number using System; class GFG{ // Function to find the // product of digits static int getProduct(int n) { int product = 1; while (n != 0) { product = product * (n % 10); n = n / 10; } return product; } // Function to find sum of all prime // factors of a given number N static int sumOfprimeFactors(int n) { int sum = 0; // Add the number of // 2s that divide n while (n % 2 == 0) { sum += 2; n = n / 2; } // N must be odd at this // point. So we can skip // one element for(int i = 3; i <= Math.Sqrt(n); i = i + 2) { // While i divides n, // add i and divide n while (n % i == 0) { sum += i; n = n / i; } } // Condition to handle the case when N // is a prime number greater than 2 if (n > 2) sum += n; return sum; } // Function to check if n // is Rhonda numberstatic bool isRhondaNum(int N) { return (10 * sumOfprimeFactors(N) == getProduct(N));} // Driver code public static void Main(String []args) { int n = 1568; if (isRhondaNum(n)) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); }} }// This code is contributed by vikas_g |
Javascript
<script>// Javascript program to check if N // is a Rhonda number // Function to find the // product of digits function getProduct( n) { let product = 1; while (n != 0) { product = product * (n % 10); n = parseInt(n / 10); } return product; } // Function to find sum of all prime // factors of a given number N function sumOfprimeFactors( n) { let sum = 0; // Add the number of // 2s that divide n while (n % 2 == 0) { sum += 2; n = parseInt(n / 2); } // N must be odd at this // point. So we can skip // one element for ( let i = 3; i <= Math.sqrt(n); i = i + 2) { // While i divides n, // add i and divide n while (n % i == 0) { sum += i; n = parseInt(n / i); } } // Condition to handle the case when N // is a prime number greater than 2 if (n > 2) sum += n; return sum; } // Function to check if n // is Rhonda number function isRhondaNum( N) { return (10 * sumOfprimeFactors(N) == getProduct(N)); } // Driver Code // Given Number n let n = 1568; if (isRhondaNum(n)) { document.write("Yes"); } else { document.write("No"); }// This code is contributed by todaysgaurav </script> |
Output:
Yes
Time Complexity: O(sqrt(N))
References: OEIS
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
