/*--.--.--.--.--.--.--.--.--.--.--.--.--*
 *  	By-Rohit Singh					 *
 *    CS , MNNIT Allahabad               *
 *		rohitmnnit1459@gmail.com         *
 *--.--.--.--.--.--.--.--.--.--.--.--.--*/

#include <iostream>
#include <cstdio>
#include <vector>
#include <map>
#include <queue>
#include <stack>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <set>
#include <limits.h>
#include <assert.h>
//#include <unordered_map> 
using namespace std;

#define mp make_pair
#define pb push_back
#define X first
#define Y second
#define null NULL
#define ll long long
#define llu unsigned long long
#define MAX 200005
#define mod 1000000007
#define inf 1e16
#define pp pair <int, int>

typedef double db;
typedef long double ldb;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair <ll, int> pli;
typedef pair <ldb, ldb> pdd;

const db PI = 3.141592653589793238;
#define abs(x) ((x)>0?(x):-(x))
//#define pop pop_back()
#define scan(x) scanf("%d",&x)
#define print(x) printf("%d\n",x)
#define scanll(x) scanf("%lld",&x)
#define scanllu(x) scanf("%llu",&x)
#define printll(x) printf("%lld\n",x)

#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define PER(i,a,b) for(int i=b;i>=a;i--)
ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
//**********************************************HAPPY CODDING*****************************************************//
int ind;
llu arr[100], n, ans, ex, gc, num;

void compute(llu n)  {
	int i, j;
	ind = arr[0] = 0;
	while (n)  {
		arr[++ind] = n%10;
		n /= 10;
	}
	for (i = 1,j = ind ; i <= j ; i++,j--)  {
		swap(arr[i], arr[j]);
	}
	REP(i,1,ind)  {
		arr[i] ^= arr[i-1];
	}
}
llu gcd(llu a, llu b)  {
	return (!a ? b : gcd(b%a, a));
}
int main()  {

	int t, rem;
	scan(t);
	while (t--)  {
		scanllu(n);
		compute(n);
		ans = 0;
		num = n;
		while (n)  {
			ind--;
			ans += (llu)(n/10)*(llu)45;
			rem = n%10;
			ex = 0;
			while (rem)  {
				ex += (arr[ind]^rem);
				rem--;
			}
			ans += ex;
			n /= 10;
		}
		gc = gcd(ans, num);
		ans /= gc;
		num /= gc;
		printf("%llu/%llu\n", ans, num);
	}
	return 0;
}