Rhonda numbers

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 digits
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
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 <= 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
bool isRhondaNum(int N)
{
    return 10 * sumOfprimeFactors(N) == getProduct(N);
}

// Driver code
int 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 Code
public 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 code
n = 1568

if (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 number
static 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