You get to the start of the trail by reading the Wikipedia article on tetration.
We know from problem 108 how the answer is formed. The trick is to generate a number of integers from the prime factors, to find which combination has the most divisors.
This was a fun one, I did try a 2nd-order brute force approach after some research and realized it would break the 1-minute rule but let it run anyway for about 10 minutes and got the answer.
Did quite some research on this but in the end extended a “classic” Hamming number generator (for 5-smooth numbers) that I found on Rosettacode.com. Bit of a cheat and runs in 2 minutes instead of the required 1.
I suck at probabilities and had to google for hints for this one.
This was fun, once I figured out the geometry. Another thing to watch out for is comparing 2 floats for equality.
The key to this is to handle the fact that 6 and 9 are equivalent. Once I’d handled that I had to eliminate duplicates. Otherwise it’s pretty straightforward.
Hey, I got an award!
Straightforward brute force, helped by the Perl Algorithm::Combinatorics module to find different permutations and so on.
Runs in about 36s with some debugging info.
Made level 5 (=125 solved problems) with this one!
This was basically the same as 173, my code needed changes to 3 lines and some hash counting code.
I’m one problem away from 125 problems solved and the next level…
On a roll today!
This wasn’t too hard once I sorted out all the fencepost errors.