## Search found 80 matches

- Thu Feb 22, 2018 10:51 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 620
- Replies:
**20** - Views:
**5849**

### Re: Problem 620

A correct configuration of gears will ALWAYS have their underlying circles as tangent to each other as described in the problem, correct? Yes. How important is the shape of the teeth? If they were more square or more rounded, would it affect the answer? What if the teeth stuck out further or lesser...

- Mon Feb 19, 2018 8:07 am
- Forum: News, Suggestions, and FAQ
- Topic: "Permanent" Posts...
- Replies:
**4** - Views:
**1226**

### Re: "Permanent" Posts...

A post is automatically archieved if it is one of the first 100 posts.kenbrooker wrote: ↑Mon Feb 19, 2018 4:31 amWhat makes a post "permanent"?

Maybe if it's within the first 200 ever?

If a post beyond that gets more than 5 kudos it has a fair chance of being archieved manually by the admins.

- Sun Jan 14, 2018 12:25 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 617
- Replies:
**6** - Views:
**2049**

### Re: Problem 617

2, 4, 2, 4, ... is indeed a different $(18,2)-MPS$ sequence than 4, 2, 4, 2, ... This follows from the problem statement Note that even though such a sequence is uniquely determined by n, e and a 0 , for most values such a sequence does not exist. and is confirmed by the examples given. The example ...

- Sun Dec 31, 2017 12:50 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 165
- Replies:
**41** - Views:
**14549**

### Re: Problem 165

Hi, CherylLynn As pointed out by traxex, it seems that you have mixed up the meaning of "distinct" intersection points with "unique" points. If you were loocking for unique points your first interpretation would have been right to completely exclude points with multiple intersections, for distinct p...

- Sat Dec 16, 2017 5:09 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 616
- Replies:
**16** - Views:
**3177**

### Re: Problem 616

It means:

n can be "converted" to

Good luck

n can be "converted" to

**every**integer m>1.Good luck

- Sat Dec 02, 2017 9:58 pm
- Forum: News, Suggestions, and FAQ
- Topic: Difficulty Rating for New Problems
- Replies:
**20** - Views:
**4070**

### Re: Difficulty Rating for New Problems

With Poohsticks Marathon the dev team was divided between medium and easy. Even for the team members being pretty well trained the subjective difficulty varies significantly. Most of us have felt stupid at same time not being able to find as good a solution (or any solution at all) for a problem tha...

- Sat Dec 02, 2017 1:38 pm
- Forum: News, Suggestions, and FAQ
- Topic: Difficulty Rating for New Problems
- Replies:
**20** - Views:
**4070**

### Re: Difficulty Rating for New Problems

Again for the same reason, 'underestimating' can occur more often. During some of the problem development discussions, I got the opinion that probability problems that are considered of medium difficulty ended up reaching max. difficulty levels ( 584 and 499 may be two examples). #499 was rated Eas...

- Fri Dec 01, 2017 1:38 am
- Forum: News, Suggestions, and FAQ
- Topic: Difficulty Rating for New Problems
- Replies:
**20** - Views:
**4070**

### Re: Difficulty Rating for New Problems

Judging the difficulty of a problem for the majority of users in advance is no easy task and the development team has often under- or overestimated it. Nevertheless, as we have repeatetly said, we try to follow a regular publishing scheme by clustering the problems into batches of 6 problems each ac...

- Sat Nov 25, 2017 12:06 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 549
- Replies:
**19** - Views:
**5746**

### Re: Problem 549

Answered via PM.

- Thu Nov 16, 2017 9:51 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 605
- Replies:
**9** - Views:
**1964**

### Re: Problem 605

So the chance the whole game is over after this two rounds is \(1/2 * 1/2 = 1/4\). And this does not depend on the number of the players. Correct. So \(P_n(2) = 1/4\) for any \(n\). False, $P_n(2)$ is the probability that the game ends after 2 rounds, or after n+2 rounds, or after 2n+2rounds ect.

- Wed Nov 15, 2017 1:04 pm
- Forum: News, Suggestions, and FAQ
- Topic: Solution thread
- Replies:
**8** - Views:
**2074**

### Re: Solution thread

Hi, Muthu we refrain from posting our own solutions on a regular basis (espcially shortly after publishing the problem), since this might take credit away from our participants solving this on their own respective taking the time to present their solutions in the forum. Moreover, we have seen that t...

- Mon Sep 11, 2017 10:55 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 609
- Replies:
**10** - Views:
**3521**

### Re: Problem 609

It means: count all $\pi$ sequences starting below n+1, that contain exactly k non-primes, e.g. p(4,1)=3.

edit: "non-primes" of course.

edit: "non-primes" of course.

- Tue Aug 08, 2017 6:46 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 041
- Replies:
**25** - Views:
**6984**

### Re: problem 41

Your number 98765431 is not pandigital according to the definition in the problem statement; having a length of 8 digits it should contain all digits from 1 to 8 exactly once.

- Tue Aug 08, 2017 11:09 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 504
- Replies:
**12** - Views:
**4533**

### Re: Problem 504

A polygon strictly contains any point that is located within the polygon and not on its boundary.

So neither count the vertices of the rectangle nor the lattice points located on the sides of the rectangle.

So neither count the vertices of the rectangle nor the lattice points located on the sides of the rectangle.

- Thu Jun 01, 2017 10:40 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 389
- Replies:
**17** - Views:
**5893**

### Re: Problem 389

The dice are 1-indexed, e.g. the numbers on the 4 sided die run from 1 to 4.

- Tue May 23, 2017 8:46 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 604
- Replies:
**3** - Views:
**2068**

### Re: Problem 604

... Especially the words 'strictly convex increasing' is confusing me.. Hi Muthu, a strictly convex increasing function is a function that is strictly convex and increasing at the same time (within the domain of definition). An increasing function is characterized by $f(a)<=f(b) \quad\text{ for all...

- Sun Apr 23, 2017 11:11 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 090
- Replies:
**46** - Views:
**15215**

### Re: Problem 090

To me it looks like your validity test is faulty. The second example in the problem description will IMHO return false with your code while testing for 09 (erroneously).

Please check this.

Please check this.

- Tue Jan 24, 2017 12:43 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 090
- Replies:
**46** - Views:
**15215**

### Re: Problem 090

Consider we are able to form all the squares with these two combinations (1) { 0, 1, 2, 6, 7, 9 } { 1, 3, 4, 5, 8, 9 } (2) { 1, 3, 4, 5, 8, 9 } { 0, 1, 2, 6, 7, 9 } the 2nd one is not the duplicate ? I guess earlier post in the forum says its duplicate. You are right, they are considered as duplica...

- Mon Jan 02, 2017 11:50 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 584
- Replies:
**9** - Views:
**3286**

### Re: Problem 584

Actually, it means within a week plus one day, since 0 days apart would mean on the same day.

So if, for example, someone would have its Birthday on some Friday, another person on Friday a week later and someone in between these two, the condition would be fulfilled.

So if, for example, someone would have its Birthday on some Friday, another person on Friday a week later and someone in between these two, the condition would be fulfilled.

- Sun Dec 25, 2016 6:54 pm
- Forum: News, Suggestions, and FAQ
- Topic: Problem difficulty cycle?
- Replies:
**3** - Views:
**1314**

### Re: Problem difficulty cycle?

Hi, vamsikal3. Regarding difficulty, we are following this publication cycle for quite a while now: medium - easy - medium - easy - medium - hard There are currently 2 more problems in the batch so the problem going to be published on Dez 31 st is a medium one, the next one after that is a hard one....