Smallest Integer Divisible by K - Problem
Given a positive integer k, you need to find the length of the smallest positive integer n such that:
nis divisible byknonly 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, so we work with remainders using modular arithmetic.
Examples of numbers containing only 1s: 1, 11, 111, 1111, 11111...
Input & Output
example_1.py — Basic Case
$
Input:
k = 1
›
Output:
1
💡 Note:
The smallest integer containing only 1s that is divisible by 1 is just '1' itself, which has length 1.
example_2.py — Small Prime
$
Input:
k = 3
›
Output:
3
💡 Note:
We check: 1%3=1, 11%3=2, 111%3=0. So '111' is the first number containing only 1s divisible by 3, with length 3.
example_3.py — Impossible Case
$
Input:
k = 2
›
Output:
-1
💡 Note:
Any number containing only 1s is odd, so it can never be divisible by 2. Return -1.
Visualization
Tap to expand
Understanding the Visualization
1
Check Feasibility
If k divisible by 2 or 5, impossible (repunits are always odd and never end in 0/5)
2
Build Remainders
Instead of huge numbers, track remainder = (remainder * 10 + 1) % k
3
Detect Success
When remainder becomes 0, we found our answer
4
Detect Cycles
If we see same remainder twice, no solution exists
Key Takeaway
🎯 Key Insight: Use modular arithmetic to track remainders instead of building massive numbers, and detect cycles to determine when no solution exists.
Time & Space Complexity
Time Complexity
O(k)
At most k different remainders possible, so we'll find answer or cycle within k iterations
✓ Linear Growth
Space Complexity
O(k)
Hash set to store seen remainders, at most k different values
✓ Linear Space
Constraints
- 1 ≤ k ≤ 105
- k is a positive integer
- Time limit: Solution must run within reasonable time for all valid inputs
💡
Explanation
AI Ready
💡 Suggestion
Tab
to accept
Esc
to dismiss
// Output will appear here after running code