1106MediumHash TableMathPigeonhole Principle

Stamping Ones Until It Divides

Tracked in this browser only

Trains the technique from

LeetCode 1015Smallest Integer Divisible by K

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A machine stamps out numbers written with the digit 1 alone: 1, then 11, then 111, and so on without end.

Return the number of digits in the shortest of these that k divides exactly, or -1 when k divides none of them.

Examples

Example 1

Input: k = 7
Output: 6

The five shortest stamps all leave a remainder, and the sixth is seven times 15873.

Example 2

Input: k = 5
Output: -1

Every stamp ends in a one, while a multiple of five has to end in a five or a zero, so no stamp can ever divide.

Example 3

Input: k = 9
Output: 9

A multiple of nine has digits adding to a multiple of nine, so nine stamps are needed, and that number is nine times 12345679.

Constraints

1 <= k <= 10^5

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def smallest_repunit_div_by_k(k: int) -> int:

Java

public int smallestRepunitDivByK(int k)

The intended approach

Time O(k) steps, each a fixed amount of arithmetic.Space O(1).

The numbers themselves race past any fixed-width integer within twenty stamps, but only their remainder against `k` matters. Each stamp turns the number into ten times itself plus one, so the remainder follows the same rule: multiply by ten, add one, reduce. That leaves at most `k` different remainders, so walking the stamps for `k` steps either lands on a remainder of zero or revisits a remainder it has already seen. A revisit means the remainders have entered a cycle that will repeat forever without ever reaching zero, which is why stopping after `k` steps loses nothing. Two cases are settled outright: every one of these numbers ends in the digit `1`, while a multiple of two ends in an even digit and a multiple of five ends in `0` or `5`. So an even `k`, or one divisible by five, divides none of them.

Pythonentry point: smallest_repunit_div_by_k

def smallest_repunit_div_by_k(k):
    if k % 2 == 0 or k % 5 == 0:
        return -1
    rest = 0
    for width in range(1, k + 1):
        rest = (rest * 10 + 1) % k
        if rest == 0:
            return width
    return -1

The obvious version

Build the numbers themselves, one digit at a time, and test each against `k` outright, stopping after `k` of them. Costs the digit count squared in arithmetic on ever-longer numbers, so it crawls once `k` reaches the thousands.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_smallest_repunit_div_by_k(k):
    value = 0
    for width in range(1, k + 1):
        value = value * 10 + 1
        if value % k == 0:
            return width
    return -1

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in