day4.cpp

#include<iostream>

using namespace std; 

/*
Suppose you have a multiplication table that is N by N. 
That is, a 2D array where the value at the i-th row and j-th column is (i + 1) * (j + 1) 
(if 0-indexed) or i * j (if 1-indexed).

Given integers N and X, write a function that returns the number of times X 
appears as a value in an N by N multiplication table.

For example, given N = 6 and X = 12, you should return 4,
 since the multiplication table looks like this:

| 1 | 2 | 3 | 4 | 5 | 6 |

| 2 | 4 | 6 | 8 | 10 | 12 |

| 3 | 6 | 9 | 12 | 15 | 18 |

| 4 | 8 | 12 | 16 | 20 | 24 |

| 5 | 10 | 15 | 20 | 25 | 30 |

| 6 | 12 | 18 | 24 | 30 | 36 |

And there are 4 12's in the table.
*/

int solution(int n, int x){
    int count = 0;
    for(int i=1;i<=n;i++){
        for (int j=1;j<=n;j++){
            if ((i*j)==x){
                count+=1;
            }
        }
    }
    return count;
}


int main(){
    cout<<"N=6 and X = 12: "<<solution(6,12)<<endl;
    return 0;
}