以3结尾的数字至少有一个具有全部的多个

Question

嗨这是Adobe面试中提出的问题.

Numbers ending in 3 have at least one multiple having all ones. 
for eg., 3 and 13 have amultiples like 111 and 111111 respectively. Given such 
a no. , find the smallest such multiple of that number. The multiple can 
exceed the range of int, long. You cannot use any complex data structure.

你能为我提供一个有效的解决方案吗？

Answer 1

得到了答案:)

int count=1, rem=1; 
while(rem)
{
 rem= (rem*10+1)%n; count++;
} 
 while(count--){ cout<<"1";}