# 1015. Smallest Integer Divisible by K

## Description

Given a positive integer k, you need to find the length of 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.

Note: n may not fit in a 64-bit signed integer.

Example 1:

Input: k = 1
Output: 1
Explanation: The smallest answer is n = 1, which has length 1.


Example 2:

Input: k = 2
Output: -1
Explanation: There is no such positive integer n divisible by 2.


Example 3:

Input: k = 3
Output: 3
Explanation: The smallest answer is n = 111, which has length 3.


Constraints:

• 1 <= k <= 105

## Solutions

• class Solution {
public int smallestRepunitDivByK(int k) {
int n = 1 % k;
for (int i = 1; i <= k; ++i) {
if (n == 0) {
return i;
}
n = (n * 10 + 1) % k;
}
return -1;
}
}

• class Solution {
public:
int smallestRepunitDivByK(int k) {
int n = 1 % k;
for (int i = 1; i <= k; ++i) {
if (n == 0) {
return i;
}
n = (n * 10 + 1) % k;
}
return -1;
}
};

• class Solution:
def smallestRepunitDivByK(self, k: int) -> int:
n = 1 % k
for i in range(1, k + 1):
if n == 0:
return i
n = (n * 10 + 1) % k
return -1


• func smallestRepunitDivByK(k int) int {
n := 1 % k
for i := 1; i <= k; i++ {
if n == 0 {
return i
}
n = (n*10 + 1) % k
}
return -1
}

• function smallestRepunitDivByK(k: number): number {
let n = 1 % k;
for (let i = 1; i <= k; ++i) {
if (n === 0) {
return i;
}
n = (n * 10 + 1) % k;
}
return -1;
}