# smallest-integer-divisible-by-k


Try it on <a href='https://leetcode.com/problems/smallest-integer-divisible-by-k'>leetcode</a>

## Description
<div class="description">
<div><p>Given a positive integer <code>k</code>, you need to find the <strong>length</strong> of the <strong>smallest</strong> positive integer <code>n</code> such that <code>n</code> is divisible by <code>k</code>, and <code>n</code> only contains the digit <code>1</code>.</p>

<p>Return <em>the <strong>length</strong> of </em><code>n</code>. If there is no such <code>n</code>, return -1.</p>

<p><strong>Note:</strong> <code>n</code> may not fit in a 64-bit signed integer.</p>

<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>

<pre><strong>Input:</strong> k = 1
<strong>Output:</strong> 1
<strong>Explanation:</strong> The smallest answer is n = 1, which has length 1.
</pre>

<p><strong>Example 2:</strong></p>

<pre><strong>Input:</strong> k = 2
<strong>Output:</strong> -1
<strong>Explanation:</strong> There is no such positive integer n divisible by 2.
</pre>

<p><strong>Example 3:</strong></p>

<pre><strong>Input:</strong> k = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> The smallest answer is n = 111, which has length 3.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= 10<sup>5</sup></code></li>
</ul>
</div>
</div>

## Solution(Python)
```Python
class Solution:
    def smallestRepunitDivByK(self, k: int) -> int:
        rem = 1
        len_n = 1

        seen = set()
        while rem % k != 0:
            n = rem * 10 + 1
            rem = n % k
            len_n += 1

            if rem in seen:
                return -1
            else:
                seen.add(rem)
        return len_n

```