## Search found 84 matches

Mon Sep 01, 2014 10:34 pm
Forum: Clarifications on Project Euler Problems
Topic: Problem 086
Replies: 77
Views: 22406

### Re: Problem 086

I hope you don't mind, but I edited out the suggestion you made as I think it gives away too much about the best way to count the number of distinct arrangements. I'm sorry for that. It was years ago I solved this problem and forget it was probably part of the problem, as it is the most natural ide...
Mon Sep 01, 2014 6:28 am
Forum: Clarifications on Project Euler Problems
Topic: Problem 086
Replies: 77
Views: 22406

### Re: Problem 086

I don't understand what was the problem with the former wording, but if after 6500 solvers you have to change the wording, I think you should use: [edited out by euler]. I'm not a native speaker, so I don't want to give you a whole sentence. And technically a permutation must consist of different ob...
Mon Aug 18, 2014 9:37 am
Forum: News, Suggestions, and FAQ
Topic: Project Euler Returns
Replies: 80
Views: 26610

### Re: Project Euler Returns

I don't know whether it is easy to solve or not, but earlier when I logged in PE new that I list only unsolved problem. It is not a big deal to press the link, but it was a good thing.
Fri Aug 15, 2014 7:38 pm
Forum: Clarifications on Project Euler Problems
Topic: Problem 016
Replies: 28
Views: 10011

### Re: Problem 016

You can send your code to me.
Fri Aug 15, 2014 6:11 pm
Forum: News, Suggestions, and FAQ
Topic: Project Euler Returns
Replies: 80
Views: 26610

### Re: Project Euler Returns

Thank you for the whole team for this work. I am very happy that even this is summer and lots of people take rest in their work, you still recovered the site.
Sun Aug 03, 2014 1:43 pm
Forum: Clarifications on Project Euler Problems
Topic: Problem 050
Replies: 42
Views: 16894

### Re: Problem 050

My old algorithm takes 1.339000 sec. It is not optimised, a C code.
I have an average laptop with I5 CPU and I'm using Windows and I don't know anything about my compiler. Sorry for the missing information.
But if you feel like to, you can send me your code or your idea, how you solved it.
Tue Jul 22, 2014 7:22 pm
Forum: Number
Topic: Brut Force?
Replies: 8
Views: 5111

### Re: Brut Force?

The easiest way to solve it took me only 10 seconds. :) The following cointains the answer, so don't read it if you don't want to see it: Google -> 1 2 3 4 to make 24 first page: https://answers.yahoo.com/question/index?qid=20080430083159AAFxkjq 24 = 6 / ( 1 - 3/4 ) Anyway for learning the basics, I...
Thu Apr 17, 2014 1:01 pm
Forum: Recreational
Topic: AD INFINITUM - MATH PROGRAMMING CONTEST
Replies: 7
Views: 5272

### Re: AD INFINITUM - MATH PROGRAMMING CONTEST

From the example it was clear what you meant, I just wanted to correct it. As mathematicans and programmers, we all should try to do our best to make precise statements.
Wed Apr 16, 2014 7:01 pm
Forum: Recreational
Topic: AD INFINITUM - MATH PROGRAMMING CONTEST
Replies: 7
Views: 5272

### Re: AD INFINITUM - MATH PROGRAMMING CONTEST

In the problem "A Chocolate Fiesta", there is a major mistake. In the note section, the set A has two elements, not three. In the structure called set, it doesn't matter, how many times an element is listed. So {2,2,3}={2,3}. That means its power set has 4 elements, not 8.
Sat Feb 22, 2014 4:18 pm
Forum: Clarifications on Project Euler Problems
Topic: Problem 035
Replies: 27
Views: 11073

### Re: Problem 035

You can send it to me, if you want.
Thu Feb 06, 2014 7:32 am
Forum: Clarifications on Project Euler Problems
Topic: Problem 064
Replies: 24
Views: 10424

### Re: Problem 064

I think problem descriptions should not contain everything. For a later problem, which had more than 1000 solutions, I spent 9 months to find out, that I don't know some obvious thing. With that knowledge, I could solve the latest problem in less than 90 minutes (I didn't wake up for the start), and...
Mon Feb 03, 2014 5:30 pm
Forum: Number
Topic: Another statement (obvious?)
Replies: 0
Views: 2317

### Re: Another statement (obvious?)

Yes, since there is a number between
a and a2,
a2 and a3,
and so on. They are in strictly increasing order, so there are infinitely many of these numbers.
Sun Feb 02, 2014 5:12 pm
Forum: Number
Topic: Statement
Replies: 1
Views: 2782

### Re: Statement

The sum of the reciprocials of natural numbers is infinity. So SUM_n=1^infinity (1/n)=infinity. If you skip the first any number of terms, the sum will stay infinity. That's why there is a border for which the sum of k reciprocials will be larger than any given number. A sketch for the proof: Make a...
Mon Jan 27, 2014 5:44 pm
Forum: Combinatorics
Topic: Placing sets in a board
Replies: 1
Views: 3203

### Re: Placing sets in a board

My solution doesn't violate any of your rules. You should change the second rule to "the meet of any two subsuts on the board has to be empty".
Mon Jan 27, 2014 5:26 pm
Forum: Combinatorics
Topic: Placing sets in a board
Replies: 1
Views: 3203

### Re: Placing sets in a board

The union will never be empty. The meet of subsets could be empty.

My solution:

Code: Select all

ABCDEFGHIJ
JABCDEFGHI
IJABCDEFGH
...

and so on, where A={1,2,3,4,5}, B={6,7,8,9,10}, ...
Sun Jan 26, 2014 5:29 pm
Forum: Programming languages
Replies: 3
Views: 3583

You can send it to me.
Wed Jan 22, 2014 7:51 am
Forum: Clarifications on Project Euler Problems
Topic: Problem 205
Replies: 27
Views: 14252

### Re: Problem 205

The order of the nine 4-sided dice doesn't matter, but they are induvudual dice, so it should be counted as different. An easier example: You have two dice (normal 6-sided), and throw both of them. If the order doesn't matter the chances of throwing (two 6's) and (one 6 + one 5) should be the same, ...
Tue Jan 14, 2014 6:13 am
Forum: Clarifications on Project Euler Problems
Topic: Problem 445
Replies: 10
Views: 4499

### Re: Problem 445

Yes, the identity map is always a retraction.
Tue Jan 07, 2014 6:53 am
Forum: Clarifications on Project Euler Problems
Topic: Problem 050
Replies: 42
Views: 16894

### Re: Problem 050

That was years ago when I solved this problem. Of course the longest chain will be at least 21 long. Since there are a lot of primes, you don't have to double your limit, just raise by 1000, or so, since their average has to be around 1000000/21, if the series cointains 21 terms. Anyway, for later p...
Mon Jan 06, 2014 8:47 pm
Forum: Clarifications on Project Euler Problems
Topic: Problem 050
Replies: 42
Views: 16894

### Re: Problem 050

First of all, the 21 terms in the sequence belongs to the largest prime below 1000. I'm almost sure, the longest sum contains more terms than 21, if the limit is 1 million. Nevertheless your upper limit is wrong, and I want to illustrate on an example. If the limit was 100 and you would have been al...