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I wrote Problem 371, https://projecteuler.net/problem=371 https://projecteuler.net/problem=371 , as a high school student in 2012! I'm so happy to have spent t
by SethTro 10mo ago
I wrote Problem 371, https://projecteuler.net/problem=371 https://projecteuler.net/problem=371 , as a high school student in 2012!
I'm so happy to have spent twenty years of my life learning math and solving problems on Project Euler and elsewhere.
- am17an 10mo agoThis is one of my favourite problems, I still remember that it has a very real edge case even though I solved it more than 10 years ago. Thank you for the problem!
- knollimar 10mo agoCan you reuse a plate with 500?
- hatthew 10mo agoThe wording seems to strongly imply no; you need two separate plates with 500 on them.
- SethTro 10mo agoI'm glad you enjoyed! It was a real game I played when driving around.
- JohnLocke4 10mo agoIs your real name also Seth? This is wholesome and hilarious
- SethTro 10mo agoYes, the other name in the problem is my sister's name :)
- vhcr 10mo agoI'm guessing if you only calculate based on the digits, the probability is going to be slightly different than the real one, because you only have a finite number of plates you can choose from.
- dpacmittal 10mo agoSounds like the birthday paradox problem. Is it?
- LanceH 10mo agoNearly, but not 8 digits of precision worth.
- dekhn 10mo agoInteresting- I ask a license plate question (when will california run out of plates in its current serialization format, based on a couple of plates observed in two different years). It's a much simpler question, though (just linear extrapolation).
- stevefan1999 10mo agoThat sounds like a combinatorial problem...alphabets from AAA to ZZZ, numbers from 000 to 999. That means one of the total sum of possible car plates is 26^3. Since we want to find pairs (x, y) that x + y = 1000. That means the total sum would also add up sum([1 for x in range(1000) for y in range(1000) if x + y == 1000])/2 since there is a symmetry. But wait, find the expected number of plates he needs to see for a win. So maybe we need to borrow something from statistics (Possion/chi-squared distribution) or queueing theory...? Edit: ah I saw the solution, it is a Markov chain.
- deleted 10mo ago[deleted]