Tcefrep Numbers
Last Updated :
23 Mar, 2021
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:
- We will find the sum of proper divisors of N
- We will find the reverse of N
- 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++
#include <bits/stdc++.h>
using namespace std;
int reverse( int num)
{
int rev_num = 0;
while (num > 0)
{
rev_num = rev_num*10 + num%10;
num = num/10;
}
return rev_num;
}
int properDivSum( int num)
{
int result = 0;
for ( int i=2; i<= sqrt (num); i++)
{
if (num%i==0)
{
if (i==(num/i))
result += i;
else
result += (i + num/i);
}
}
return (result + 1);
}
bool isTcefrep( int n)
{
return properDivSum(n) == reverse(n);
}
int main()
{
int N = 6;
if (isTcefrep(N))
cout << "Yes" ;
else
cout << "No" ;
return 0;
}
|
Java
class GFG{
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;
}
static int properDivSum( int num)
{
int result = 0 ;
for ( int i = 2 ; i<= Math.sqrt(num); i++)
{
if (num % i == 0 )
{
if (i == (num / i))
result += i;
else
result += (i + num / i);
}
}
return (result + 1 );
}
static boolean isTcefrep( int n)
{
return properDivSum(n) == reverse(n);
}
public static void main(String[] args)
{
int N = 6 ;
if (isTcefrep(N))
System.out.print( "Yes" );
else
System.out.print( "No" );
}
}
|
Python3
import math
def reverse(num):
rev_num = 0
while (num > 0 ):
rev_num = rev_num * 10 + num % 10
num = num / / 10
return rev_num
def properDivSum(num):
result = 0
for i in range ( 2 , ( int )(math.sqrt(num)) + 1 ):
if (num % i = = 0 ):
if (i = = (num / / i)):
result + = i
else :
result + = (i + num / i)
return (result + 1 )
def isTcefrep(n):
return properDivSum(n) = = reverse(n);
N = 6
if (isTcefrep(N)):
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
class GFG{
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;
}
static int properDivSum( int num)
{
int result = 0;
for ( int i = 2; i<= Math.Sqrt(num); i++)
{
if (num % i == 0)
{
if (i == (num / i))
result += i;
else
result += (i + num / i);
}
}
return (result + 1);
}
static bool isTcefrep( int n)
{
return properDivSum(n) == reverse(n);
}
public static void Main()
{
int N = 6;
if (isTcefrep(N))
Console.Write( "Yes" );
else
Console.Write( "No" );
}
}
|
Javascript
<script>
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 properDivSum( num) {
let result = 0;
for ( i = 2; i <= Math.sqrt(num); i++) {
if (num % i == 0) {
if (i == (num / i))
result += i;
else
result += (i + num / i);
}
}
return (result + 1);
}
function isTcefrep( n) {
return properDivSum(n) == reverse(n);
}
let N = 6;
if (isTcefrep(N))
document.write( "Yes" );
else
document.write( "No" );
</script>
|
Time Complexity: O(n^2)
Reference: http://www.numbersaplenty.com/set/tcefrep_number/
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