Given a number N, the task is to check if N is an Duffinian Number or not. If N is an Duffinian Number then print “Yes” else print “No”.
Duffinian Number is a composite numbers N such that it is relatively prime to the sum of divisors of N.
Examples:
Input: N = 35
Output: Yes
Explanation:
Sum of divisors of 35 = 1 + 5 + 7 + 35 = 48,
and 48 is relatively prime to 35.
Input: N = 28
Output: No
Explanation:
Sum of divisors of 28 = 1 + 2 + 4 + 7 + 14 + 28 = 56,
and 56 is not relatively prime to 28.
Approach: The idea is to find the sum of factors of the number N. If the gcd of N and the sum of factors of N are relatively prime then the given number N is a Duffinian Number, else N is not a Duffinian Number.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to calculate the sum of all // divisors of a given number int divSum( int n)
{ // Sum of divisors
int result = 0;
// Find all divisors of num
for ( int i = 2; i <= sqrt (n); i++) {
// if 'i' is divisor of 'n'
if (n % i == 0) {
// If both divisors are same
// then add it once
if (i == (n / i))
result += i;
else
result += (i + n / i);
}
}
// Add 1 and n to result as above
// loop considers proper divisors
// greater than 1.
return (result + n + 1);
} // Function to check if n is an // Duffinian number bool isDuffinian( int n)
{ // Calculate the sum of divisors
int sumDivisors = divSum(n);
// If number is prime return false
if (sumDivisors == n + 1)
return false ;
// Find the gcd of n and sum of
// divisors of n
int hcf = __gcd(n, sumDivisors);
// Returns true if N and sumDivisors
// are relatively prime
return hcf == 1;
} // Driver Code int main()
{ // Given Number
int n = 36;
// Function Call
if (isDuffinian(n))
cout << "Yes" ;
else
cout << "No" ;
return 0;
} |
// Java program for the above approach class GFG{
// Recursive function to return // gcd of a and b static int gcd( int a, int b)
{ if (b == 0 )
return a;
return gcd(b, a % b);
} // Function to calculate the sum of // all divisors of a given number static int divSum( int n)
{ // Sum of divisors
int result = 0 ;
// Find all divisors of num
for ( int i = 2 ; i <= Math.sqrt(n); i++)
{
// if 'i' is divisor of 'n'
if (n % i == 0 )
{
// If both divisors are same
// then add it once
if (i == (n / i))
result += i;
else
result += (i + n / i);
}
}
// Add 1 and n to result as above
// loop considers proper divisors
// greater than 1.
return (result + n + 1 );
} // Function to check if n is an // Duffinian number static boolean isDuffinian( int n)
{ // Calculate the sum of divisors
int sumDivisors = divSum(n);
// If number is prime return false
if (sumDivisors == n + 1 )
return false ;
// Find the gcd of n and sum of
// divisors of n
int hcf = gcd(n, sumDivisors);
// Returns true if N and sumDivisors
// are relatively prime
return hcf == 1 ;
} // Driver code public static void main(String[] args)
{ // Given Number
int n = 36 ;
// Function Call
if (isDuffinian(n))
{
System.out.println( "Yes" );
}
else
{
System.out.println( "No" );
}
} } // This code is contributed by shubham |
# Python3 program for the above approach import math
# Function to calculate the sum of all # divisors of a given number def divSum(n):
# Sum of divisors
result = 0
# Find all divisors of num
for i in range ( 2 , int (math.sqrt(n)) + 1 ):
# If 'i' is divisor of 'n'
if (n % i = = 0 ):
# If both divisors are same
# then add it once
if (i = = (n / / i)):
result + = i
else :
result + = (i + n / i)
# Add 1 and n to result as above
# loop considers proper divisors
# greater than 1.
return (result + n + 1 )
# Function to check if n is an # Duffinian number def isDuffinian(n):
# Calculate the sum of divisors
sumDivisors = int (divSum(n))
# If number is prime return false
if (sumDivisors = = n + 1 ):
return False
# Find the gcd of n and sum of
# divisors of n
hcf = math.gcd(n, sumDivisors)
# Returns true if N and sumDivisors
# are relatively prime
return hcf = = 1
# Driver Code # Given number n = 36
# Function call if (isDuffinian(n)):
print ( "Yes" )
else :
print ( "No" )
# This code is contributed by sanjoy_62 |
// C# program for the above approach using System;
class GFG{
// Recursive function to return // gcd of a and b static int gcd( int a, int b)
{ if (b == 0)
return a;
return gcd(b, a % b);
} // Function to calculate the sum of // all divisors of a given number static int divSum( int n)
{ // Sum of divisors
int result = 0;
// Find all divisors of num
for ( int i = 2; i <= Math.Sqrt(n); i++)
{
// If 'i' is divisor of 'n'
if (n % i == 0)
{
// If both divisors are same
// then add it once
if (i == (n / i))
result += i;
else
result += (i + n / i);
}
}
// Add 1 and n to result as above
// loop considers proper divisors
// greater than 1.
return (result + n + 1);
} // Function to check if n is an // Duffinian number static bool isDuffinian( int n)
{ // Calculate the sum of divisors
int sumDivisors = divSum(n);
// If number is prime return false
if (sumDivisors == n + 1)
return false ;
// Find the gcd of n and sum of
// divisors of n
int hcf = gcd(n, sumDivisors);
// Returns true if N and sumDivisors
// are relatively prime
return hcf == 1;
} // Driver code public static void Main( string [] args)
{ // Given number
int n = 36;
// Function call
if (isDuffinian(n))
{
Console.Write( "Yes" );
}
else
{
Console.Write( "No" );
}
} } // This code is contributed by rock_cool |
<script> // Javascript program for the above approach // Recursive function to return
// gcd of a and b
function gcd( a , b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function to calculate the sum of
// all divisors of a given number
function divSum( n) {
// Sum of divisors
let result = 0;
// Find all divisors of num
for (let i = 2; i <= Math.sqrt(n); i++) {
// if 'i' is divisor of 'n'
if (n % i == 0) {
// If both divisors are same
// then add it once
if (i == (n / i))
result += i;
else
result += (i + n / i);
}
}
// Add 1 and n to result as above
// loop considers proper divisors
// greater than 1.
return (result + n + 1);
}
// Function to check if n is an
// Duffinian number
function isDuffinian( n) {
// Calculate the sum of divisors
let sumDivisors = divSum(n);
// If number is prime return false
if (sumDivisors == n + 1)
return false ;
// Find the gcd of n and sum of
// divisors of n
let hcf = gcd(n, sumDivisors);
// Returns true if N and sumDivisors
// are relatively prime
return hcf == 1;
}
// Driver code
// Given Number
let n = 36;
// Function Call
if (isDuffinian(n)) {
document.write( "Yes" );
} else {
document.write( "No" );
}
// This code contributed by Rajput-Ji </script> |
Yes
Time Complexity: O(1)