link

Code：

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
#define int long long
const int N = 1e5 + 10, M = 1e5;
int c;
int qmul(int a,int p,int mod){
    int res=0;
    while(p){
        if(p&1)res=(res+a)%mod;
        a=(a+a)%mod;
        p>>=1;
    }
    return res;
}//慢速乘，用于解决快速幂中的两个1e10相乘爆ll问题
int gcd(int a, int b) {
	return b ? gcd(b, a%b) : a;
}
int qmi(int a, int p, int mod) {
	int res = 1;
	while (p) {
		if (p & 1)res = qmul(res,a,mod)%mod;
		a = qmul(a,a,mod)%mod;
		p >>= 1;
	}
	return res;
}
bool check(int x) {
	if (qmi(10, x, c) == 1)return 1;
	return 0;
}
signed main()
{
	int L;
	int id = 0;
	while (cin >> L, L) {
		printf("Case %lld: ", ++id);
		c = 9 * L / gcd(L, 8);
		if (gcd(c, 10) != 1)cout << 0 << endl;
		else {
			int phi = c;
			int tc = c;
			for (int i = 2; i <= M; i++) {
				int cnt = 0;
				while (tc%i == 0) {
					cnt++;
					tc /= i;
				}
				if(cnt)phi = phi * (i - 1) / i;
			}//求欧拉函数
			if(tc!=1)phi = phi /tc *(tc-1);//先除后乘，不然又爆ll了...
			
			int res = phi;
			for (int i = 1; i*i<=phi; i++) {
				if (phi%i == 0 ){
				    if(check(i))res=min(res,i);
				    if(check(phi/i))res=min(res,phi/i); //求约数
				    
				}
				
			}
		    cout<<res<<endl;
 
		}
	}
}

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