#include <iostream>

#include <cstdio>

#include <cstring>

#include <string>

#include <algorithm>

#include <map>

#include <queue>

#include <cmath>

using namespace std;

int main()

{

    int n, a[], first = ;

    while (scanf("%d", &n) != EOF,n) {

        if (first) {

            first = ;

        }

        else {

            printf("\n");

        }                                //以上是两组数据之间输出空行的技巧

        int num1 = , num2 = ;

        int side =  / n;           //这里稍微降低了一下复杂度

        int flag1 = ;

        for (num1 = ; num1 <= side; num1++) {

            int flag = ;

            num2 = num1 * n;

            a[] = num2 / ;

            a[] = num2 /  % ;

            a[] = num2 /  % ;

            a[] = num2 /  % ;

            a[] = num2 % ;

            a[] = num1 / ;

            a[] = num1 /  % ;

            a[] = num1 /  % ;

            a[] = num1 /  % ;

            a[] = num1 % ;

            for (int i = ; i < ; i++) {

                for (int j = ; j < ; j++) {

                    if (j != i && a[i] == a[j]) {            //这里判断10位数是否有重复的方法

                        flag = ;

                    }

                }

            }

            if (flag) {

                for (int i = ; i < ; i++)cout << a[i]; cout << " / ";                        //注意这里"/"和"="左右两边都有空格

                for (int i = ; i < ; i++)cout << a[i]; cout << " = " << n << endl;

                flag1 = ;

            }

        }

        if (flag1 == ) {

            printf("There are no solutions for %d.\n", n);

        }

    }

    return ;

}