Search found 84 matches

by TheEvil
Mon Jan 07, 2013 2:50 pm
Forum: Clarifications on Project Euler Problems
Topic: Problem 126
Replies: 28
Views: 9414

Re: Problem 126

If you read the thread carefully, you will see that others had the same result. There are more than 3 layers possible, which was a typical problem.
by TheEvil
Mon Jan 07, 2013 10:16 am
Forum: Clarifications on Project Euler Problems
Topic: Problem 097
Replies: 15
Views: 4230

Re: Problem 097

If you were asked the last two digit of this number, your answer would almost be good. But you have to calculate
(28433%100) × 27830457%20 + 1
and of course at the end, the last two digit of that number. Anyway the problem is solvable without bigintegers (as all the others I have done yet).
by TheEvil
Thu Jan 03, 2013 10:47 am
Forum: Clarifications on Project Euler Problems
Topic: Problem 233
Replies: 36
Views: 15004

Re: Problem 233

When I was trying to solve this problem, I could find a formula for f(N), but I spent a day with finding that little mistake, like everyone else. If you have something for 'primitive' N-s, you should try to find something for general N-s. And after that you won't have double counting problems.
by TheEvil
Mon Dec 17, 2012 10:55 pm
Forum: Clarifications on Project Euler Problems
Topic: Problem 086
Replies: 77
Views: 18783

Re: Problem 086

rigel wrote:6,7,1 -> 6*6 + (7+1)*(7+1) = 10*10
I think there are more misunderstandings, but this is not the shortest route for this cuboid. There is one little shorter for this cuboid, but it is not integer length.
by TheEvil
Mon Dec 17, 2012 5:07 pm
Forum: Clarifications on Project Euler Problems
Topic: Problem 190
Replies: 6
Views: 2959

Re: Problem 190

The point is, that P_m should not be integer. If your answer (a=32/27) is correct, then you have to take the greatest integer which is less or equal to a, in this case, you should type 1 as a correct answer.
by TheEvil
Sun Dec 16, 2012 11:39 pm
Forum: Clarifications on Project Euler Problems
Topic: Problem 054
Replies: 74
Views: 23863

Re: Problem 054

If you want, you can send me a list of your results, and I will give you the first wrong row. Or if you are misunderstanding something, it can be easier, if I make it clear in Hungarian.
by TheEvil
Thu Dec 13, 2012 6:15 pm
Forum: Clarifications on Project Euler Problems
Topic: Problem 135
Replies: 29
Views: 10523

Re: Problem 135

however I have a question : how could n have a 'least' value? doesn't that term implies there is also a 'most' value and there are more than one possible 'n's for an arithmetic sequence? If I remember well, in most of the cases least means the smallest number. By this I mean: let us fix an arbitrar...
by TheEvil
Wed Dec 05, 2012 9:47 pm
Forum: Clarifications on Project Euler Problems
Topic: Problem 078
Replies: 16
Views: 6582

Re: Problem 078

As a simple brute force could show with increasing n, the value of p(n) increases so much that it is hard to imagine, it would fit in the long long type. And it is the case, so if you would like to calculate it you will need bigger type, but it is solvable without that (as almost all of the PE probl...
by TheEvil
Mon Nov 05, 2012 8:15 pm
Forum: Clarifications on Project Euler Problems
Topic: Problem 103
Replies: 38
Views: 10789

Re: Problem 103

rouge6789 wrote:Bonsoir.
Pour 4, vous indiquez que l'optimum est 3,5,6,7 soit un total de 21
Mon algorithme trouve : 2,3,4 8 pour un total de 17 !!
Je vois pas mon erreur ?
Cordialement.
Condition ii.:
B={2,3}
C={8}
|B|=2 > 1=|C|
although 2+3<8, which is contradiction

Sorry, for not speaking French.
by TheEvil
Thu Oct 25, 2012 8:29 pm
Forum: Clarifications on Project Euler Problems
Topic: Problem 072
Replies: 36
Views: 9576

Re: Problem 072

I'm not sure I have understood your problem right, but if you want to generate all the solutions, without a supercomputer, you will fail, since those answers are not "too" far from the correct one, and you can see, that there are more than 10^11 different fractions for the limit one million. There a...
by TheEvil
Mon Oct 08, 2012 10:45 pm
Forum: Clarifications on Project Euler Problems
Topic: Problem 144
Replies: 20
Views: 8265

Re: Problem 144

The first one is incorrect, so I would say, the others are bad too (I haven't tested). This won't be your problem probably, but look for the counterm to calculate reflections correctly (+/-1 are common mistake, at least for me). Could someone clarify me on this? I wonder if I have precision problem ...
by TheEvil
Tue Aug 28, 2012 9:14 pm
Forum: Recreational
Topic: A problem for every age
Replies: 20
Views: 9106

Re: A problem for every age

Nice version. It is actually a little harder, I like it. I am seriously thinking of giving this problem in one of my test papers at the university to see, how fresh minds can solve problems like this. :)
by TheEvil
Tue Aug 28, 2012 10:40 am
Forum: Recreational
Topic: A problem for every age
Replies: 20
Views: 9106

Re: A problem for every age

euler wrote:What a lovely little puzzle. Thanks for sharing it.
Happy to hear that. Of course your suggestion makes it a little more interesting. Even with TeX one can do it in no more than 5 minutes, if you need the code I can send it to you (that's even faster :D ).
by TheEvil
Sun Aug 26, 2012 11:30 am
Forum: Recreational
Topic: A problem for every age
Replies: 20
Views: 9106

Re: A problem for every age

The problem is that most of the adults do not dare to connect anything. After I realised that I have to start, it was very easy, as you have said.
by TheEvil
Sat Aug 25, 2012 8:55 pm
Forum: News, Suggestions, and FAQ
Topic: Functional requests
Replies: 194
Views: 45372

Re: Functional requests

lsmith946 wrote:Rather trivial again but it would be good to see which awardsI have earned rather than just the number I have earned.
You can see all of them on your Progress page between the Problems page and Friends page.
by TheEvil
Sat Aug 25, 2012 8:03 pm
Forum: Recreational
Topic: A problem for every age
Replies: 20
Views: 9106

A problem for every age

I want to start with the background of the problem. First of all, when I heard it years before I was said to be that it is an easy problem, since an average kindergarten pupil (age 4 to 6) can solve it in about 15 seconds. It took me a half day to solve it with a little help. After I solved it I rea...
by TheEvil
Sat Aug 25, 2012 10:38 am
Forum: Clarifications on Project Euler Problems
Topic: Problem 054
Replies: 74
Views: 23863

Re: Problem 054

If both hands are three-of-a-kind, first compare the three-of-a-kind; if equal, .... In such a case, someone is definitely cheating if playing with a 52-card deck!!! :( :shock: :lol: Actually, that's not necessarily true. Imagine this: Player 1: Ad Ts Player 2: Td 7h Board: 5s Th 8d 3s Tc Both play...
by TheEvil
Fri Aug 24, 2012 5:01 pm
Forum: Clarifications on Project Euler Problems
Topic: Problem 054
Replies: 74
Views: 23863

Re: Problem 054

Yes it is true. But lot's of other possibilities don't occur. If you could find that, you must be able to solve the problem.
by TheEvil
Thu Aug 23, 2012 10:00 am
Forum: Clarifications on Project Euler Problems
Topic: Problem 023
Replies: 56
Views: 20214

Re: Problem 023

Hi, I have a question about this one: In the problem it states that "As 12 is the smallest abundant number...." . Is this really true ? Let's take the number 11. Its proper divisors are 1 and 11 (since it's prime). 1 + 11 = 12, which means it's abundant. Actually all prime numbers are abundant. Are...
by TheEvil
Wed Aug 22, 2012 4:50 pm
Forum: Clarifications on Project Euler Problems
Topic: Problem 145
Replies: 18
Views: 6296

Re: Problem 145

The problem says that leading zeros are not allowed but I don't think that it's clear how to treat numbers ending in zeros. Should they be completely discarded since reverse(n) would contain leading zeros? Or should the leading zeros in reverse(n) be trimmed? And if so then which digits would corre...