Looking at more Project Euler problems.

I thought problem 41 was going to be tough, but there are only 9! = 362880 nine-digit numbers to consider, and even fewer 8-digit numbers to consider, etc, so it was just a matter of checking for the first prime permutation.

Problem 42 was straightforward. I made a list of triangle numbers up to the maximum word value, then filtered the word values and counted.

For problem 43, I started by taking all multiples of 17 under 1000, filtering out any with repeated digits. Adding the 4th digit made the set 7 times bigger, but removing those that did not have the divisible by 13 property made the set about half the size. Iterate for the remaining digits.

I couldn't find a reasonable way to solve problem 44. It involves finding 4 pentagonal numbers, p_i, p_j, p_k, p_l, where p_i ≤ p_j < p_k < p_l, and p_i = p_k - p_j, and p_j + p_k = p_l, where p_i is minimized. For a given i, there's an upper bound for j, such that p_i ≥ p_j+1 - p_j, and the upper bound for j is O(i²), which means a brute-force search over i gets extremely slow. However, since I didn't have any better ideas, I searched i from 1 to about 350 or so without finding a solution, and gave up.

For problem 45, I made a list of hexagonal numbers, filtered out those that weren't pentagonal or triangle, and took the 3rd number in the resulting list, the first two being 1 and 40755. It would have been slightly slower to make a list of pentagonal or triangle numbers, and filter out those that weren't the other types.

For problem 46, I iterated over non-prime odd numbers, building up a list of primes and doubled squares less than it, stopping when none of the primes could be added to one of the doubled squares to get that number.

In problem 47, I only had to check every 4th number until one had at least 4 factors, then check the previous one, the one before that, and the one before that. If any didn't have at least 4 factors, I could jump forward 4 from that.

For problem 48, using mod 10¹⁰ arithmetic means not having to add gigantic numbers, while still getting the last 10 digits.

For problem 49, I took all the 4 digit primes and grouped the ones that were permutations of each other together, then searched each group for a qualifying triplet.

For problem 50, I found the longest prime sum starting with 2, as well as the list of primes after the last term of the sum. Then, I subtracted 2 and added the next largest prime, looking for the longest sum starting with 3, and iterating until subtracting the next smallest prime and adding enough of the next largest primes to make the sum have more terms than the current longest sum resulted in a sum more than a million.