D
Description
If an integer is not divisible by 2 or 5, some multiple of that number in decimal notation is a sequence of only a digit. Now you are given the number and the only allowable digit, you should report the number of digits of such multiple.
For example you have to find a multiple of 3 which contains only 1's. Then the result is 3 because is 111 (3-digit) divisible by 3. Similarly if you are finding some multiple of 7 which contains only 3's then, the result is 6, because 333333 is divisible by 7.
Input
Input starts with an integer T (≤ 300), denoting the number of test cases.
Each case will contain two integers n (0 < n ≤ 106 and n will not be divisible by 2 or 5) and the allowable digit (1 ≤ digit ≤ 9).
Output
For each case, print the case number and the number of digits of such multiple. If several solutions are there; report the minimum one.
Sample Input
3
3 1
7 3
9901 1
Sample Output
1 #include<cstdio> 2 #include<cmath> 3 #include<cstring> 4 int main() 5 { 6 int t,i,j,ans,num,n,digit; 7 scanf("%d",&t); 8 for(i = 1;i <= t;i++) 9 { 10 scanf("%d %d",&n,&digit); 11 ans = digit ; 12 j = 1; 13 num =1; 14 while(1) 15 { 16 if(ans%n!=0) 17 { 18 ans=ans%n*10+digit; 19 j++; 20 num++; 21 } 22 if(ans%n==0) 23 { 24 break; 25 } 26 } 27 printf("Case %d: %d\n",i,num); 28 } 29 return 0; 30 }
Case 1: 3
Case 2: 6
Case 3: 12