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++
// C++ implementation to check if N // is a Tcefrep number#include <bits/stdc++.h> using namespace std; // Iterative function to // reverse digits of numint 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 Codeint main() { // Given Number N int N = 6; // Function Call if (isTcefrep(N)) cout << "Yes"; else cout << "No"; return 0; } |
Java
// Java program for above approachclass GFG{ // Iterative function to // reverse digits of numstatic 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 numberimport math# Iterative function to # reverse digits of numdef 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 approachusing System;class GFG{ // Iterative function to // reverse digits of numstatic 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/
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