Advent of Code F# – Day 20

Yan Cui

I help clients go faster for less using serverless technologies.

The source code for this post (both Part 1 and Part 2) is available here and you can click here to see my solutions for the other Advent of Code challenges.

Description for today’s challenge is here.


Once you understood what this challenge is really asking the solution becomes really simple:


If you read the question carefully, what it’s really saying is:

For every house number N, only elves whose number M that is a divisor of N will visit this house and deliver M * 10 gifts. Find the smallest N where Sum(M * 10) >= 29000000.

So our first task is to find all divisors for any given number N, and since each divisor has a opposite (as in, for a divisor M, N/M is also a divisor) so you only need to iterate up to Sqrt(N).

And since my input is such a high number, we can safely start from a relatively high number for N too (100K in this case) and use Seq.unfold to generate an infinite sequence of numbers.


Part 2

Part 2 adds a small twist to proceedings – only divisors M that are <= N/50 will be considered, and the multiplier is now 11 instead of 10.


Notice that since we’re not looking for ‘all’ divisors anymore, so I also took the liberty to rename the function to findDivisors so it more accurately describes what it does now.


Whenever you’re ready, here are 4 ways I can help you:

  1. If you want a one-stop shop to help you quickly level up your serverless skills, you should check out my Production-Ready Serverless workshop. Over 20 AWS Heroes & Community Builders have passed through this workshop, plus 1000+ students from the likes of AWS, LEGO, Booking, HBO and Siemens.
  2. If you want to learn how to test serverless applications without all the pain and hassle, you should check out my latest course, Testing Serverless Architectures.
  3. If you’re a manager or founder and want to help your team move faster and build better software, then check out my consulting services.
  4. If you just want to hang out, talk serverless, or ask for help, then you should join my FREE Community.


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