Advent of Code F# – Day 20

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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:

day20_01

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.

day20_02

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.

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