The kth Factor of n
A medium-tier problem at 70% community acceptance, tagged with Math, Number Theory. Reported in interviews at Expedia and 2 others.
The kth Factor of n is a deceptively simple problem that trips up candidates who overthink it. You need to find the kth smallest factor of a given number n. Expedia, IBM, and Salesforce have all asked it. The acceptance rate sits at 70%, but that's misleading, most who fail don't understand the constraint properly or implement a solution that's slower than it needs to be. If this hits your live OA and the mathematical pattern doesn't click immediately, StealthCoder surfaces a working solution in seconds, invisible to the proctor.
Companies that ask "The kth Factor of n"
The kth Factor of n is the kind of problem that decides whether you pass. StealthCoder reads the problem on screen and surfaces a working solution in under 2 seconds. Invisible to screen share. The proctor sees nothing. Built by an engineer who got tired of watching his cohort grind for six months and still get filtered at the OA stage.
Get StealthCoderThe trap is assuming you need complex number theory. You don't. You iterate from 1 to n, check divisibility, count factors, and return the kth one. The real edge comes from recognizing you only need to check up to the square root of n, then work backwards to find the actual kth factor efficiently. Most candidates either brute-force all the way to n (slow) or fail to handle the symmetry of factors correctly. Math and Number Theory topics here are straightforward, no prime factorization tricks, no modular arithmetic. This is fundamentally an iteration problem that looks like it belongs in a harder category. If you blank during the assessment and can't optimize fast enough, StealthCoder runs the correct loop for you while the proctor sees nothing.
Pattern tags
You know the problem.
Make sure you actually pass it.
The kth Factor of n recycles across companies for a reason. It's medium-tier, and most candidates blank under the timer. StealthCoder is the hedge: an AI overlay invisible during screen share. It reads the problem and surfaces a working solution in under 2 seconds. Built by an engineer who got tired of watching his cohort grind for six months and still get filtered at the OA stage. Works on HackerRank, CodeSignal, CoderPad, and Karat.
The kth Factor of n interview FAQ
Is this really a MEDIUM problem?+
Yes, but the difficulty is psychological, not algorithmic. The factor-finding logic is elementary. The edge case handling, when k exceeds the total number of factors, trips people up. And optimization beyond naive iteration isn't obvious on first read, which is why the acceptance rate is 70%, not 90%.
Do I need to know advanced number theory?+
No. This problem is tagged Number Theory but doesn't require it. You iterate, count divisors, and return the kth one. Basic divisibility checks are all you need. The number theory framing is somewhat misleading.
What's the trick to making it fast?+
Only iterate up to sqrt(n). For each divisor i you find, you also know n/i is a divisor. Use this symmetry to build your factor list without checking every number up to n. This cuts runtime dramatically on large n.
What happens if k is larger than the total number of factors?+
Return -1 (or whatever the problem specifies). Most candidates hardcode their loop to n and don't check if k is valid beforehand. Read the constraints carefully to see what the problem expects in this edge case.
Will Expedia, IBM, and Salesforce actually ask this?+
Yes, all three are reported askers. It's a mid-level assessment problem that filters for clean implementation and basic optimization. Expect it in online coding assessments, not as a follow-up or whiteboard question.
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