#include <bits/stdc++.h>

using namespace std;
 /** 1/k = 1/x + 1/y
            When x and y are equal, they are both 2k.

            1/k = 1/(2k) + 1/(2k)

            if x is larger than 2k, then y must be smaller than 2k.
            Since y can't be k, we only need to check y in [k+1, 2k].
            But how to compute x?
            1/x = 1/k - 1/y
                = (y-k) / (k*y)

            So if y-k divides k*y, then x = (k*y) / (y-k).
        */
int main()
{
int k;
while(cin>>k)
{
    int c=0;
    vector<pair<int,int>>v;
    for(int i=k+1;i<=2*k;i++)
    {
        if(i*k%(i-k)==0)
        {
            c++;
            v.push_back(make_pair((i*k)/(i-k),i));
        }

    }
    cout<<c<<endl;
    for(int i=0;i<v.size();i++){
    cout<<"1/"<<k<<" = 1/"<<v[i].first<<" + "<<"1/"<<v[i].second<<endl;;

    }

}
return 0;
}