Formatted question description: https://leetcode.ca/all/1015.html

# 1015. Smallest Integer Divisible by K

## Level

Medium

## Description

Given a positive integer `K`

, you need find the **smallest** positive integer `N`

such that `N`

is divisible by `K`

, and `N`

only contains the digit 1.

Return the length of `N`

. If there is no such `N`

, return -1.

Example 1:

**Input:** 1

**Output:** 1

**Explanation:** The smallest answer is N = 1, which has length 1.

**Example 2:**

**Input:** 2

**Output:** -1

**Explanation:** There is no such positive integer N divisible by 2.

**Example 3:**

**Input:** 3

**Output:** 3

**Explanation:** The smallest answer is N = 111, which has length 3.

**Note:**

`1 <= K <= 10^5`

## Solution

If `K`

is a multiple of 2 or 5, then any number divisible by `K`

can’t end with 1, so return `-1`

.

Otherwise, each time append a 1 to the current number (which is initially 0) and check whether the new number is divisible by `K`

. Use modulo operation to avoid overflow.

```
class Solution {
public int smallestRepunitDivByK(int K) {
if (K % 2 == 0 || K % 5 == 0)
return -1;
int value = 1;
int length = 1;
while (value % K != 0) {
value = value % K * 10 + 1;
length++;
}
return length;
}
}
```