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

# 1015. Smallest Integer Divisible by K

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;
}
}

• // OJ: https://leetcode.com/problems/smallest-integer-divisible-by-k/
// Time: O(K)
// Space: O(K)
class Solution {
public:
int smallestRepunitDivByK(int k) {
unordered_set<int> s;
for (int n = 1 % k, len = 1; true; n = (n * 10 + 1) % k, ++len) {
if (n == 0) return len;
if (s.count(n)) return -1;
s.insert(n);
}
}
};

• # 1015. Smallest Integer Divisible by K
# https://leetcode.com/problems/smallest-integer-divisible-by-k/

class Solution:
def smallestRepunitDivByK(self, k: int) -> int:
if k % 2 == 0 and k % 5 == 0: return -1
mod_set = set()
prev = 0

for i in range(1, k + 1):
prev = (prev * 10 + 1) % k
if prev == 0: return i
if prev in mod_set: return -1

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;
}