Count pairs of numbers from 1 to N with Product divisible by their Sum in C++
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
The approach used in the below program is as follows
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.
Example
Live Demo
#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;
}Output
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
Published on 02-Dec-2020 15:47:57
