Amicable numbers are two different numbers so related that the sum of the proper divisors of each is equal to the other number. (A proper divisor of a number is a positive factor of that number other than the number itself.
Examples:
Input : x = 220, y = 284 Output : Yes Proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110. Sum of these is 284. Proper divisors of 284 are 1, 2, 4, 71 and 142 with sum 220. Input : 1 2 Output :No
The logic is very simple. We compare sum of the proper divisors of both numbers and compare sum for one number with other number.
C++
// CPP program to check if two numbers are // Amicable or not. #include <bits/stdc++.h> using namespace std;
// Function to calculate sum of all // proper divisors of a given number int divSum( int n)
{ // Sum of divisors
int result = 0;
// find all divisors which divides '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 else add
// both
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 + 1);
} // Returns true if x and y are Amicable // else false. bool areAmicable( int x, int y)
{ if (divSum(x) != y)
return false ;
return (divSum(y) == x);
} int main() {
int x = 220, y = 284;
if (areAmicable(x, y))
cout << "Yes" ;
else
cout << "No" ;
return 0;
} |
Java
// JAVA program to check if two numbers are // Amicable or not. import java.io.*;
class GFG
{ // Function to calculate sum of all
// proper divisors of a given number
static int divSum( int n)
{
// Sum of divisors
int result = 0 ;
// find all divisors which divides '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 else add
// both
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 + 1 );
}
// Returns true if x and y are Amicable
// else false.
static boolean areAmicable( int x, int y)
{
if (divSum(x) != y)
return false ;
return (divSum(y) == x);
}
public static void main (String[] args)
{
int x = 220 , y = 284 ;
if (areAmicable(x, y))
System.out.println( "Yes" );
else
System.out.println( "No" );
}
} // This code is contributed by vt_m. |
Python3
# Python program to check # if two numbers are # Amicable or not. import math
# def to calculate sum # of all proper divisors # of a given number def divSum(n) :
# Sum of divisors
result = 0
# find all divisors
# which divides '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 else add
# both
if (i = = int (n / i)) :
result = result + i
else :
result = result +
(i + int (n / i))
# Add 1 and n to result
# as above loop considers
# proper divisors greater
# than 1.
return (result + 1 )
# Returns true if x and y # are Amicable else false. def areAmicable(x, y) :
if (divSum(x) ! = y) :
return False
return (divSum(y) = = x)
# Driver Code x = 220
y = 284
if (areAmicable(x, y)) :
print ( "Yes" )
else :
print ( "No" )
# This code is contributed by # Manish Shaw(manishshaw1) |
C#
// C# program to check if two numbers are // Amicable or not. using System;
class GFG
{ // Function to calculate sum of all
// proper divisors of a given number
static int divSum( int n)
{
// Sum of divisors
int result = 0;
// find all divisors which divides '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 else add
// both
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 + 1);
}
// Returns true if x and y are Amicable
// else false.
static bool areAmicable( int x, int y)
{
if (divSum(x) != y)
return false ;
return (divSum(y) == x);
}
public static void Main ()
{
int x = 220, y = 284;
if (areAmicable(x, y))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
} // This code is contributed by vt_m. |
PHP
<?php // PHP program to check if two // numbers are Amicable or not. // Function to calculate sum of all // proper divisors of a given number function divSum( $n )
{ // Sum of divisors
$result = 0;
// find all divisors
// which divides 'num'
for ( $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 else add
// both
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 + 1);
} // Returns true if x and y // are Amicable else false. function areAmicable( $x , $y )
{ if (divSum( $x ) != $y )
return false;
return (divSum( $y ) == $x );
} // Driver Code
$x = 220;
$y = 284;
if (areAmicable( $x , $y ))
echo "Yes" ;
else
echo "No" ;
// This code is contributed by anuj_67. ?> |
Javascript
<script> // Javascript program to check if two
// numbers are Amicable or not. // Function to calculate sum of all // proper divisors of a given number function divSum(n)
{ // Sum of divisors
let result = 0;
// find all divisors
// which divides '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 else add
// both
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 + 1);
} // Returns true if x and y // are Amicable else false. function areAmicable(x, y)
{ if (divSum(x) != y)
return false ;
return (divSum(y) == x);
} // Driver Code
let x = 220;
let y = 284;
if (areAmicable(x, y))
document.write( "Yes" );
else
document.write( "No" );
// This code is contributed by _saurabh_jaiswal. </script> |
Output:
Yes
Time Complexity: O(?n)
Auxiliary Space: O(1)
Please suggest if someone has a better solution which is more efficient in terms of space and time.