# Check if the given two numbers are friendly pair or not

• Difficulty Level : Medium
• Last Updated : 13 Sep, 2021

Given two positive integer N, M. The task is to check if N and M are friendly pair or not.

In number theory, friendly pairs are two numbers with a common abundancy index, the ratio between the sum of divisors of a number and the number itself i.e ?(n)/n. S o, two number n and m are friendly number if
?(n)/n = ?(m)/m.
where ?(n) is the sum of divisors of n.

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Examples:

```Input : n = 6, m = 28
Output : Yes
Explanation:
Divisor of 6 are 1, 2, 3, 6.
Divisor of 28 are 1, 2, 4, 7, 14, 28.
Sum of divisor of 6 and 28 are 12 and 56
respectively. Abundancy index of 6 and 28
are 2. So they are friendly pair.

Input : n = 18, m = 26
Output : No```

The idea is to find the sum of divisor of n and m. And to check if the abundancy index of n and m, we will find the Greatest Common Divisors of n and m with their sum of divisors. And check if the reduced form of abundancy index of n and m are equal by checking if their numerator and denominator are equal or not. To find the reduced form, we will divide numerator and denominator by GCD.
Below is the implementation of above idea :

## C++

 `// Check if the given two number``// are friendly pair or not.``#include ``using` `namespace` `std;` `// Returns sum of all factors of n.``int` `sumofFactors(``int` `n)``{` `    ``// Traversing through all prime factors.``    ``int` `res = 1;``    ``for` `(``int` `i = 2; i <= ``sqrt``(n); i++) {` `        ``int` `count = 0, curr_sum = 1;``        ``int` `curr_term = 1;``        ``while` `(n % i == 0) {``            ``count++;` `            ``// THE BELOW STATEMENT MAKES``            ``// IT BETTER THAN ABOVE METHOD``            ``// AS WE REDUCE VALUE OF n.``            ``n = n / i;` `            ``curr_term *= i;``            ``curr_sum += curr_term;``        ``}` `        ``res *= curr_sum;``    ``}` `    ``// This condition is to handle``    ``// the case when n is a prime``    ``// number greater than 2.``    ``if` `(n >= 2)``        ``res *= (1 + n);` `    ``return` `res;``}` `// Function to return gcd of a and b``int` `gcd(``int` `a, ``int` `b)``{``    ``if` `(a == 0)``        ``return` `b;``    ``return` `gcd(b % a, a);``}` `// Function to check if the given two``// number are friendly pair or not.``bool` `checkFriendly(``int` `n, ``int` `m)``{``    ``// Finding the sum of factors of n and m``    ``int` `sumFactors_n = sumofFactors(n);``    ``int` `sumFactors_m = sumofFactors(m);` `    ``// finding gcd of n and sum of its factors.``    ``int` `gcd_n = gcd(n, sumFactors_n);` `    ``// finding gcd of m and sum of its factors.``    ``int` `gcd_m = gcd(m, sumFactors_m);` `    ``// checking is numerator and denominator of``    ``// abundancy index of both number are equal``    ``// or not.``    ``if` `(n / gcd_n == m / gcd_m &&``        ``sumFactors_n / gcd_n == sumFactors_m / gcd_m)``        ``return` `true``;` `    ``else``        ``return` `false``;``}` `// Driver code``int` `main()``{``    ``int` `n = 6, m = 28;``    ``checkFriendly(n, m) ? (cout << ``"Yes\n"``) :``                          ``(cout << ``"No\n"``);``    ``return` `0;``}`

## Java

 `// Java code to check if the given two number``// are friendly pair or not.``class` `GFG {``    ` `    ``// Returns sum of all factors of n.``    ``static` `int` `sumofFactors(``int` `n)``    ``{``    ` `        ``// Traversing through all prime factors.``        ``int` `res = ``1``;``        ``for` `(``int` `i = ``2``; i <= Math.sqrt(n); i++) {``    ` `            ``int` `count = ``0``, curr_sum = ``1``;``            ``int` `curr_term = ``1``;``            ``while` `(n % i == ``0``) {``                ``count++;``    ` `                ``// THE BELOW STATEMENT MAKES``                ``// IT BETTER THAN ABOVE METHOD``                ``// AS WE REDUCE VALUE OF n.``                ``n = n / i;``    ` `                ``curr_term *= i;``                ``curr_sum += curr_term;``            ``}``    ` `            ``res *= curr_sum;``        ``}``    ` `        ``// This condition is to handle``        ``// the case when n is a prime``        ``// number greater than 2.``        ``if` `(n >= ``2``)``            ``res *= (``1` `+ n);``    ` `        ``return` `res;``    ``}``    ` `    ``// Function to return gcd of a and b``    ``static` `int` `gcd(``int` `a, ``int` `b)``    ``{``        ``if` `(a == ``0``)``            ``return` `b;``            ` `        ``return` `gcd(b % a, a);``    ``}``    ` `    ``// Function to check if the given two``    ``// number are friendly pair or not.``    ``static` `boolean` `checkFriendly(``int` `n, ``int` `m)``    ``{``        ``// Finding the sum of factors of n and m``        ``int` `sumFactors_n = sumofFactors(n);``        ``int` `sumFactors_m = sumofFactors(m);``    ` `        ``// finding gcd of n and sum of its factors.``        ``int` `gcd_n = gcd(n, sumFactors_n);``    ` `        ``// finding gcd of m and sum of its factors.``        ``int` `gcd_m = gcd(m, sumFactors_m);``    ` `        ``// checking is numerator and denominator of``        ``// abundancy index of both number are equal``        ``// or not.``        ``if` `(n / gcd_n == m / gcd_m &&``            ``sumFactors_n / gcd_n == sumFactors_m / gcd_m)``            ``return` `true``;``    ` `        ``else``            ``return` `false``;``    ``}``    ` `    ``//driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `n = ``6``, m = ``28``;``        ` `        ``if``(checkFriendly(n, m))``            ``System.out.print(``"Yes\n"``);``        ``else``            ``System.out.print(``"No\n"``);``    ``}``}` `// This code is contributed by Anant Agarwal.`

## Python3

 `# Check if the given two number``# are friendly pair or not.``import` `math` `# Returns sum of all factors of n.``def` `sumofFactors(n):` `    ``# Traversing through all prime factors.``    ``res ``=` `1``    ``for` `i ``in` `range``(``2``, ``int``(math.sqrt(n)) ``+` `1``):` `        ``count ``=` `0``; curr_sum ``=` `1``; curr_term ``=` `1``        ``while` `(n ``%` `i ``=``=` `0``):``            ``count ``+``=` `1` `            ``# THE BELOW STATEMENT MAKES``            ``# IT BETTER THAN ABOVE METHOD``            ``# AS WE REDUCE VALUE OF n.``            ``n ``=` `n ``/``/` `i` `            ``curr_term ``*``=` `i``            ``curr_sum ``+``=` `curr_term``        ` `        ``res ``*``=` `curr_sum``    ` `    ``# This condition is to handle``    ``# the case when n is a prime``    ``# number greater than 2.``    ``if` `(n >``=` `2``):``        ``res ``*``=` `(``1` `+` `n)` `    ``return` `res` `# Function to return gcd of a and b``def` `gcd(a, b):` `    ``if` `(a ``=``=` `0``):``        ``return` `b``    ``return` `gcd(b ``%` `a, a)` `# Function to check if the given two``# number are friendly pair or not.``def` `checkFriendly(n, m):` `    ``# Finding the sum of factors of n and m``    ``sumFactors_n ``=` `sumofFactors(n)``    ``sumFactors_m ``=` `sumofFactors(m)` `    ``# Finding gcd of n and sum of its factors.``    ``gcd_n ``=` `gcd(n, sumFactors_n)` `    ``# Finding gcd of m and sum of its factors.``    ``gcd_m ``=` `gcd(m, sumFactors_m)` `    ``# checking is numerator and denominator ``    ``# of abundancy index of both number are``    ``# equal or not.``    ``if` `(n ``/``/` `gcd_n ``=``=` `m ``/``/` `gcd_m ``and``        ``sumFactors_n ``/``/` `gcd_n ``=``=` `sumFactors_m ``/``/` `gcd_m):``        ``return` `True` `    ``else``:``        ``return` `False` `# Driver code``n ``=` `6``; m ``=` `28``if``(checkFriendly(n, m)):``    ``print``(``"Yes"``)``else``:``    ``print``(``"No"``)``    ` `# This code is contributed by Anant Agarwal.`

## C#

 `// C# code to check if the``// given two number are``// friendly  pair or not``using` `System;` `class` `GFG {``    ` `    ``// Returns sum of all``    ``//factors of n.``    ``static` `int` `sumofFactors(``int` `n)``    ``{``    ` `        ``// Traversing through all``        ``// prime factors.``        ``int` `res = 1;``        ``for` `(``int` `i = 2; i <= Math.Sqrt(n); i++)``        ``{``    ` `            ``int` `count = 0, curr_sum = 1;``            ``int` `curr_term = 1;``            ``while` `(n % i == 0)``            ``{``                ``count++;``    ` `                ``// THE BELOW STATEMENT MAKES``                ``// IT BETTER THAN ABOVE METHOD``                ``// AS WE REDUCE VALUE OF n.``                ``n = n / i;``    ` `                ``curr_term *= i;``                ``curr_sum += curr_term;``            ``}``    ` `            ``res *= curr_sum;``        ``}``    ` `        ``// This condition is to handle``        ``// the case when n is a prime``        ``// number greater than 2.``        ``if` `(n >= 2)``            ``res *= (1 + n);``    ` `        ``return` `res;``    ``}``    ` `    ``// Function to return gcd``    ``// of a and b``    ``static` `int` `gcd(``int` `a, ``int` `b)``    ``{``        ``if` `(a == 0)``            ``return` `b;``            ` `        ``return` `gcd(b % a, a);``    ``}``    ` `    ``// Function to check if the``    ``// given two number are``    ``// friendly pair or not``    ``static` `bool` `checkFriendly(``int` `n, ``int` `m)``    ``{``        ``// Finding the sum of factors``        ``// of n and m``        ``int` `sumFactors_n = sumofFactors(n);``        ``int` `sumFactors_m = sumofFactors(m);``    ` `        ``// finding gcd of n and``        ``// sum of its factors.``        ``int` `gcd_n = gcd(n, sumFactors_n);``    ` `        ``// finding gcd of m and sum``        ``// of its factors.``        ``int` `gcd_m = gcd(m, sumFactors_m);``    ` `        ``// checking is numerator and``        ``// denominator of abundancy``        ``// index of both number are``        ``// equal or not``        ``if` `(n / gcd_n == m / gcd_m &&``            ``sumFactors_n / gcd_n ==``            ``sumFactors_m / gcd_m)``            ``return` `true``;``    ` `        ``else``            ``return` `false``;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main (String[] args)``    ``{``        ``int` `n = 6, m = 28;``        ` `        ``if``(checkFriendly(n, m))``            ``Console.Write(``"Yes\n"``);``        ``else``            ``Console.Write(``"No\n"``);``    ``}``}` `// This code is contributed by parshar...`

## PHP

 `= 2)``        ``\$res` `*= (1 + ``\$n``);` `    ``return` `\$res``;``}` `// Function to return``// gcd of a and b``function` `gcd(``\$a``, ``\$b``)``{``    ``if` `(``\$a` `== 0)``        ``return` `\$b``;``    ``return` `gcd(``\$b` `% ``\$a``, ``\$a``);``}` `// Function to check if the given two``// number are friendly pair or not.``function` `checkFriendly(``\$n``, ``\$m``)``{``    ` `    ``// Finding the sum of``    ``// factors of n and m``    ``\$sumFactors_n` `= sumofFactors(``\$n``);``    ``\$sumFactors_m` `= sumofFactors(``\$m``);` `    ``// finding gcd of n and``    ``// sum of its factors.``    ``\$gcd_n` `= gcd(``\$n``, ``\$sumFactors_n``);` `    ``// finding gcd of m and``    ``// sum of its factors.``    ``\$gcd_m` `= gcd(``\$m``, ``\$sumFactors_m``);` `    ``// checking is numerator``    ``// and denominator of``    ``// abundancy index of``    ``// both number are equal``    ``// or not.``    ``if` `(``\$n` `/ ``\$gcd_n` `== ``\$m` `/ ``\$gcd_m` `and``        ``\$sumFactors_n` `/ ``\$gcd_n` `==``        ``\$sumFactors_m` `/ ``\$gcd_m``)``        ``return` `true;` `    ``else``        ``return` `false;``}` `    ``// Driver code``    ``\$n` `= 6;``    ``\$m` `= 28;``    ``if``(checkFriendly(``\$n``, ``\$m``))``        ``echo` `"Yes"` `;``    ``else``        ``echo` `"No"``;` `// This code is contributed by anuj_67.``?>`

## Javascript

 ``

Output

`Yes`

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