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We are given a number N. The goal is to find the pairs of numbers from 1 to N such that the product of pairs is equal to the sum of pairs.

Let us understand with examples.

**Input** − N=11

**Output** − Count of pairs of no. from 1 to N with Product divisible by their Sum are − 1

**Explanation** − Numbers 3 and 6 have product 18 and their sum 9 fully divides 18.

**Input** − N=30

**Output** − Count of pairs of no. from 1 to N with Product divisible by their Sum are − 12

**Explanation** − Pairs are − (3, 6), (4,12), (5, 20), (6, 12), (6, 30), (8, 24), (9, 18), (10, 15), (12, 24), (15, 30), (20, 30), (21, 28)

Count of pairs − 12

We will traverse from 1 to N using FOR loop twice. For every, i search for j such that the product of (i,j) is divisible by sum i+j. Increment count for suc i,j pairs such that i!=j.

Take a number N as input.

Function Sum_N(N) takes N and returns the count of pairs such that the product of numbers is divisible by sum of numbers.

Traverse from i=1 to i<N.

Travers from j=i+1 to j<=N.

Take the initial count as 0.

For each i and j calculate temp= (i*j)%(i+j).

If temp is 0 then sum fully divides the product. Increment count.

After the end of all iterations, the count will have a total number of such pairs.

Return count as result.

#include <bits/stdc++.h> using namespace std; int Sum_N(int N){ int count = 0; for (int i = 1; i < N; i++){ for (int j = i + 1; j <= N; j++){ int temp = (j * i) % (j + i); if (!temp){ count++; } } } return count; } int main(){ int N = 20; cout<<"Count of pairs of numbers from 1 to N with Product divisible by their Sum are: "<<Sum_N(N); return 0; }

If we run the above code it will generate the following output −

Count of pairs of numbers from 1 to N with Product divisible by their Sum are: 6

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