1015. Smallest Integer Divisible by K
- let the number be
n.
- to keep 1 in no. keep adding multiplying like
n = n * 10 + 1.
- to avoid memory limit to be exceeded keep modulating
n with k.
- if we encounter any multiple of
k we immediately will get 0.
Code implementation
```cpp
class Solution {
public:
int smallestRepunitDivByK(int k) {
if (k % 2 == 0 or k % 5 == 0) return -1;
long long int num = 1;
int ans = 1;
if (num % k == 0) return ans;
while (num) {
num = num * 10 + 1;
num %= k;
ans ++;
}
return ans;
}
};
```