Tcefrep Numbers

Tcefrep Number a number N such that reverse(n) = sum of the proper divisors of N.

6, 498906, 20671542, 41673714…. 

Check if N is a Tcefrep number

Given a number N, the task is to check if N is a Tcefrep Number or not. If N is an Tcefrep Number then print “Yes” else print “No”.
Examples:

Input: N = 498906 
Output: Yes 
Explanation:
proper divisors of 498906 are 1, 2, 3, 6, 9, 18, 27, 54, 
9239, 18478, 27717, 55434, 83151, 166302, 249453, 
which sum to 609894, the reverse of 498906
Input: N = 120 
Output: No 

Approach:

1. We will find the sum of proper divisors of N
2. We will find the reverse of N
3. Then we will check if sum of proper divisors of N is equal to reverse of N or not, if equal then print “Yes” else print “No”.

Below is the implementation of the above approach:

C++

// C++ implementation to check if N
// is a Tcefrep number
#include <bits/stdc++.h>
using namespace std;

// Iterative function to
// reverse digits of num
int reverse(int num)
{
    int rev_num = 0;
    while(num > 0)
    {
        rev_num = rev_num*10 + num%10;
        num = num/10;
    }
    return rev_num;
}

// Function to calculate sum of
// all proper divisors
// num --> given natural number
int properDivSum(int num)
{
    // Final result of summation of divisors
    int result = 0;

    // find all divisors which divides 'num'
    for (int i=2; i<=sqrt(num); i++)
    {
        // if 'i' is divisor of 'num'
        if (num%i==0)
        {
            // if both divisors are same then add
            // it only once else add both
            if (i==(num/i))
                result += i;
            else
                result += (i + num/i);
        }
    }

    // Add 1 to the result as 1 is also a divisor
    return (result + 1);
}

bool isTcefrep(int n)
{
    return properDivSum(n) == reverse(n);
}

// Driver Code
int main()
{
    // Given Number N
    int N = 6;

    // Function Call
    if (isTcefrep(N))
        cout << "Yes";
    else
        cout << "No";
    return 0;

}

Java

// Java program for above approach
class GFG{

// Iterative function to
// reverse digits of num
static int reverse(int num)
{
    int rev_num = 0;
    while(num > 0)
    {
        rev_num = rev_num * 10 + num % 10;
        num = num / 10;
    }
    return rev_num;
}

// Function to calculate sum of
// all proper divisors
// num --> given natural number
static int properDivSum(int num)
{
    // Final result of summation of divisors
    int result = 0;

    // find all divisors which divides 'num'
    for (int i = 2; i<= Math.sqrt(num); i++)
    {
        // if 'i' is divisor of 'num'
        if (num % i == 0)
        {
            // if both divisors are same then add
            // it only once else add both
            if (i == (num / i))
                result += i;
            else
                result += (i + num / i);
        }
    }

    // Add 1 to the result as 1
    // is also a divisor
    return (result + 1);
}

static boolean isTcefrep(int n)
{
    return properDivSum(n) == reverse(n);
}

// Driver Code
public static void main(String[] args)
{
    int N = 6;

    // Function Call
    if (isTcefrep(N))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}

// This code is contributed by Pratima Pandey

Python3

# Python3 implementation to check if N
# is a Tcefrep number
import math

# Iterative function to
# reverse digits of num
def reverse(num):
    rev_num = 0
    while(num > 0):
        rev_num = rev_num * 10 + num % 10
        num = num // 10

    return rev_num

# Function to calculate sum of
# all proper divisors
# num --> given natural number
def properDivSum(num):

    # Final result of summation of divisors
    result = 0

    # find all divisors which divides 'num'
    for i in range(2, (int)(math.sqrt(num)) + 1):

        # if 'i' is divisor of 'num'
        if (num % i == 0):

            # if both divisors are same then add
            # it only once else add both
            if (i == (num // i)):
                result += i
            else:
                result += (i + num / i)

    # Add 1 to the result as 1 is also a divisor
    return (result + 1)

def isTcefrep(n):
    return properDivSum(n) == reverse(n);

# Driver Code

# Given Number N
N = 6

# Function Call
if(isTcefrep(N)):
    print("Yes")
else:
    print("No")

# This code is contributed by Sanjit Prasad

C#

// C# program for above approach
using System;
class GFG{

// Iterative function to
// reverse digits of num
static int reverse(int num)
{
    int rev_num = 0;
    while(num > 0)
    {
        rev_num = rev_num * 10 + num % 10;
        num = num / 10;
    }
    return rev_num;
}

// Function to calculate sum of
// all proper divisors
// num --> given natural number
static int properDivSum(int num)
{
    // Final result of summation of divisors
    int result = 0;

    // find all divisors which divides 'num'
    for (int i = 2; i<= Math.Sqrt(num); i++)
    {
        // if 'i' is divisor of 'num'
        if (num % i == 0)
        {
            // if both divisors are same then add
            // it only once else add both
            if (i == (num / i))
                result += i;
            else
                result += (i + num / i);
        }
    }

    // Add 1 to the result as 1
    // is also a divisor
    return (result + 1);
}

static bool isTcefrep(int n)
{
    return properDivSum(n) == reverse(n);
}

// Driver Code
public static void Main()
{
    int N = 6;

    // Function Call
    if (isTcefrep(N))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}

// This code is contributed by Nidhi_Biet

Javascript

<script>

// Javascript program for above approach


    // Iterative function to
    // reverse digits of num
    function reverse( num) {
        let rev_num = 0;
        while (num > 0) {
            rev_num = rev_num * 10 + num % 10;
            num = parseInt(num / 10);
        }
        return rev_num;
    }

    // Function to calculate sum of
    // all proper divisors
    // num --> given natural number
    function properDivSum( num) {
        // Final result of summation of divisors
        let result = 0;

        // find all divisors which divides 'num'
        for ( i = 2; i <= Math.sqrt(num); i++) {
            // if 'i' is divisor of 'num'
            if (num % i == 0) {
                // if both divisors are same then add
                // it only once else add both
                if (i == (num / i))
                    result += i;
                else
                    result += (i + num / i);
            }
        }

        // Add 1 to the result as 1
        // is also a divisor
        return (result + 1);
    }

    function isTcefrep( n) {
        return properDivSum(n) == reverse(n);
    }

    // Driver Code

        let N = 6;

        // Function Call
        if (isTcefrep(N))
            document.write("Yes");
        else
            document.write("No");

// This code contributed by gauravrajput1

</script>

Output:

Yes

Time Complexity: O(n^2)
Reference: http://www.numbersaplenty.com/set/tcefrep_number/