Norway


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

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

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  and 56
respectively. Abundancy index of 6 and 28 
are 2. So they are friendly .

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 C++ implementation of this approach:

// Check if the given two number 
// are friendly pair or not.
#include <bits/stdc++.h>
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);

    // findig 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 << "Yesn") : 
                          (cout << "Non");
    return 0;
}

Output

Yes


- avatar - Check if the given two number are friendly pair or not

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