## Prime Power Triples

### May 22, 2020

Today’s exercise is Project Euler Problem 87:

The smallest number expressible as the sum of a prime square, prime cube, and prime fourth power is 28. In fact, there are exactly four numbers below fifty that can be expressed in such a way:

28 = 2^{2}+ 2^{3}+ 2^{4}33 = 3^{2}+ 2^{3}+ 2^{4}49 = 5^{2}+ 2^{3}+ 2^{4}47 = 2^{2}+ 3^{3}+ 2^{4}How many numbers below fifty million can be expressed as the sum of a prime square, prime cube, and prime fourth power?

Your task is to solve Project Euler 87; in the spirit of Project Euler, show only your code but not the solution. When you are finished, you are welcome to read

or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

Nifty little drill. Here is my take on it using Julia 1.4.1: https://pastebin.com/DVjwfbAx

The original approach I took was way too sluggish, so I ended up using a slicker one. There is also a verbose option for viewing the specific combinations: main(n, true), where n is the maximum integer to explore (in this case 50 000 000). The default for this parameter is false.

Glad the blog is back! Cheers

Easy to solve by brute force, once you realise that there are not that many prime powers to look at.

Welcome back! Hope all is well with everybody.

Not sure I like those multiple calls to primerange etc. Here’s another Python solution that generates a single list of prime numbers. Takes a couple of seconds for the Euler problem (here we just count up to a million):

Very glad to hear you are well and that things are settling down somewhat.

Here’s my solution in Python, using a standard Sieve of Eratosthenes and precalculating the required prime powers. Runs in about half a second on my desktop.

A solution in (not necessarily well-written) Icon. Also, note that prime-power triples aren’t unique; for example, 210 = 11^2 + 2^3 + 3^4 = 2^2 + 5^3 + 3^4. Code that doesn’t handle duplicates is producing incorrect counts.

Here’s a solution in Python.

Here’s a refactored version of my last solution.

Here’s another refactor to make it shorter, at the cost of readability.