Project Euler Forum

If there is a solution along similar lines to the one for $\lfloor \sqrt{k} \rfloor$ then it's a polynomial in $n$ and $\lfloor \sqrt{n} \rfloor$. Asymptotically the solution is $\Theta(n^2)$, so there are only 9 coefficients to determine. Pick 9 values of $n$, perform Gaussian elimination to get th...

Nothing wrong with cleverness, but I do think that one of the qualifications for a problem to make it to PE is that it should be a true combination of math and programming, so a problem shouldn't be solvable without a computer doing at least some computation. The compromise position is to include t...

It would be awesome if it was possible for me to add a script which could check the solutions to the puzzles in the builds which would fail if incorrect. For that use case you don't want an API: you want a local file with test cases. Otherwise your build can randomly fail due either to temporary ne...

Who's the target audience? I've answered a few PE questions on my phone on the bus or in the café in my coffee break, and the only app I needed was qPython3.

there is inevitably some spoilage each of the five per-product spoilage rates was worse (higher) for 'B' than for 'A' These quotes from the problem statement exclude solutions with no spoilage for one of more products. I think that is less than crystal clear. It's a perfectly reasonable (and, in ge...

I think one useful tool is simply to classify. A large chunk of problems are fundamentally about prime factorisations of integers from 1 to N: i.e. number theory. (They can be further classified: e.g. problems which rely on efficient counting of primes up to N). A smaller chunk of problems are about...

The question states The rules do not place any restriction on the number of occurrences of M, so all integers have a valid representation. but doesn't explain how to get negative integers, and I got the tick with an answer which assumes that negative integers don't have a valid representation. I thi...

There's an error in the problem statement. It currently says

the denominator is a power of 3 i.e. $b=3^k, k>0$.

It should say

the denominator is a power of 3 other than 1, i.e. $b=3^k, k>0$.

(The other possible correction, $>$ to $\ge$, doesn't match up with the test cases).

For those interested in knowing, the minimal information script has been re-implemented. I have updated the first post to reflect any changes that may have taken place. There seems to be one change, possibly more recent than that, which isn't reflected in the first post: when logged in, I'm seeing ...

On another site, I recently tackled a question asked about an "idle" computer game which boiled down to the expected time to reach the absorbing state in a Markov process. As I didn't know the standard approach (I have now learnt it, so don't feel obliged to explain it :wink:), I reasoned as follows...

Aha. Thanks.

I was expecting to see either Problem 253 or at least a countdown to it today. Have I missed an announcement? Is something up?

Another obvious extension is the regular "solids" of infinite volume. There are three, with respectively 6 triangles, 4 squares, or 3 hexagons meeting at each vertex, and tiling the surface of an infinitely large sphere.

tiny wrote:The text says: whose perimeters do not exceed 1.000.000.000

Does this mean, that for every triangle the perimeter is less than 1.000.000.000?

Almost. It means less than or equal to 1.000.000.000.

I leave it as an exercise for the reader to prove that the un ordered partition of n health between r gladiators is (n+r-1)Cn. I think there is a typo there. Yes. For n=160, r=16, I find the number of ordered partitions to be much larger : 175 C 160 =1823828930470225269045 and the number of unorder...

Huh, that looks interesting. How did you calculate that number? It's the ordered partition* of 160 health between 16 gladiators. I leave it as an exercise for the reader to prove that the unordered partition of n health between r gladiators is (n+r-1)Cn. * The partition into n distinguishing the n....

I make it a Markov model with 236236542585120 states. Fortunately the transition matrix is sparse

#!/bin/sh tr A-Z a-z </usr/share/dict/british-english | sort -u >/tmp/wordlist cat /tmp/wordlist | tr a-z phxdtl%brjzfvn%aqiyeum%csk | grep -v "%" | sort >/tmp/translated cat /tmp/wordlist /tmp/translated | sort | uniq -d Not that many. There are a handful of 5-letter ones: buffy hulls bulls huffy ...

The spoiler seems to be broken, so I've unspoilered this. Sorry. There's an easily proved lower bound of 2.52; it's achieved by the following 38 sets. I used a brute force search considering only coin values up to 100. The puzzle is so closely related to Golomb rulers that I thought offsetting one o...