By considering all cuboid rooms with integer dimensions, up to a maximum size ...

## Search found 261 matches

- Fri Jan 22, 2010 7:20 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 086
- Replies:
**77** - Views:
**18946**

### Re: Problem 086

Good spot ! Wording amended to:

- Tue Jan 12, 2010 3:08 pm
- Forum: Combinatorics
- Topic: Coin Flipping
- Replies:
**20** - Views:
**9558**

### Re: Coin Flipping

Assuming that the two strings have the same number of letters, say k letters selected from an alphabet of m letters (where m ge 2), and that "-->" means "limit as k rarr infin ", I am afraid I cannot see how the condition E( N B ) / E( N A ) --> 0 could be satisfied: We know (see my previous post) t...

- Fri Jan 08, 2010 11:14 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 004
- Replies:
**83** - Views:
**16844**

### Re: Problem 004

Hi theslimzmassive, Please do not post in this forum the answers or any spoilers for any problem; this forum is open to everyone (whether they have solved a particular problem or not), so spoilers are not allowed. As a result, I snipped the correct answer from your post. As far as I can tell, the an...

- Tue Dec 29, 2009 1:13 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 162
- Replies:
**37** - Views:
**12990**

### Re: Problem 162

@friol : Please do not start a new topic if one already exists for the given problem.

- Sat Dec 26, 2009 6:50 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 270
- Replies:
**13** - Views:
**4500**

### Re: Problem 270

Yep! So far, so good

- Sat Dec 26, 2009 6:04 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 270
- Replies:
**13** - Views:
**4500**

### Re: problem 270

Nope. The numbers you give are too high I'm afraid.

- Fri Dec 18, 2009 9:02 pm
- Forum: News, Suggestions, and FAQ
- Topic: Accesibility of new problems
- Replies:
**7** - Views:
**4069**

### Re: Accesibility of new problems

Since the main PE site is still inaccessible from many parts of the world (due to DNS servers slowly updating their tables), it was thought best to postpone the release for problem #269 for 12 hours (to Saturday 19th Dec, 9 am GMT). Hopefully, everything will be working smoothly by then. Our apologi...

- Sun Nov 29, 2009 12:51 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 209
- Replies:
**20** - Views:
**7547**

### Re: Problem 209

I see now that I need to generate 64 bit strings, not 6, but here's an example from 6. Do I have this right? One of the columns (A through something more than P) has a value of 001001. Let's call that column Q. If Q in the second table evaluates to 010010 and I AND the two strings, I get 000000. Re...

- Sat Nov 28, 2009 6:55 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 209
- Replies:
**20** - Views:
**7547**

### Re: Problem 209

1. Is it necessary that I understand each column, or can I just take the logic "as is"? You just take it "as is"; as I've already mentioned before, most of the possible truth-tables do not a have a clear interpretation: that's why they don't have a name. 2. To make a 3 input table T(x, y, z), do I ...

- Sat Nov 28, 2009 4:32 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 209
- Replies:
**20** - Views:
**7547**

### Re: Problem 209

A six-input binary truth table has indeed 2 6 =64 rows. However, there are 2 64 different six-input truth tables. As an example, consider 2-input binary truth tables: each one of them has 4 rows, but there are 2 4 =16 such truth-tables, as shown below: . x y : A B C D E F G H I J K L M N O P --- - -...

- Tue Nov 24, 2009 1:33 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 151
- Replies:
**27** - Views:
**12668**

### Re: Problem 151

No, I'm afraid that the values you posted are wrong :( During each week, and excluding the first and last batch: - the probability that he finds a single sheet in the envelope exactly once is 0.3100 (rounded to 4 dec.pl.) - the probability that he finds a single sheet in the envelope exactly twice i...

- Sat Nov 21, 2009 5:52 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 223 & 224
- Replies:
**22** - Views:
**9618**

### Re: Problem 223 & 224

Yes, viv_ban, your answer for

**Problem 223**(View Problem) and perimeter ≤10000, seems to be correct.- Thu Nov 19, 2009 7:08 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 263
- Replies:
**27** - Views:
**9172**

### Re: Problem 263

It's a sum, but it's not a sum of distinct divisors; it doesmrtollefson wrote:why is 1 + 1 + 1 + 1 not a sum?

**not**have to be a sum of only one or two divisors, but the divisors making up the sum must be

**distinct**(i.e. different from each other).

- Wed Nov 18, 2009 7:10 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 264
- Replies:
**18** - Views:
**5696**

### Re: Problem 264

Yes (for your result) and yes (we know, since we were the first ones to solve them)

- Wed Nov 18, 2009 6:54 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 264
- Replies:
**18** - Views:
**5696**

### Re: Problem 264

You are missing the two largest pairs, I believe.

- Wed Nov 18, 2009 8:37 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 264
- Replies:
**18** - Views:
**5696**

### Re: Problem 264

This time however you seem to have received a nice green check Well done!Rodinio wrote:So I'm used to waiting a bit longer for that next red cross

- Wed Nov 18, 2009 8:31 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 264
- Replies:
**18** - Views:
**5696**

### Re: Problem 264

@zwuupeape:

Your result for max. perim = 500 is about half of what I find.

You seem to have missed two triangles with a perimeter<209

[ A(32,9), B(-4,-33), C(-23, 24) perim=172.4093 and its symmetric ]

plus a few triangles with perimeter>210, e.g.

A(45,10), B(-10,-45), C(-30,35) perim=239.3008

Your result for max. perim = 500 is about half of what I find.

You seem to have missed two triangles with a perimeter<209

[ A(32,9), B(-4,-33), C(-23, 24) perim=172.4093 and its symmetric ]

plus a few triangles with perimeter>210, e.g.

A(45,10), B(-10,-45), C(-30,35) perim=239.3008

- Tue Nov 17, 2009 6:02 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 264
- Replies:
**18** - Views:
**5696**

### Re: Problem 264

@zwuupeape: Nope, I'm afraid your results are too low (and not close to half the actual values). :( More specifically, your result for 2500 is quite a bit larger than half and your result for 10000 quite a bit less than half. So it seems that as you increase the perimeter limit, you miss more and mo...

- Tue Nov 17, 2009 2:03 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 049
- Replies:
**30** - Views:
**10576**

### Re: Problem 049

According to the 49 , you should be looking for three 4-digit numbers. Then you concatenate them and submit the 12-digits as your answer. E.g. if your three numbers were 1111, 2222 and 3333 you would submit 111122223333 as your answer. Also, before starting a new topic for a problem, please make sur...

- Sun Nov 15, 2009 5:48 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 264
- Replies:
**18** - Views:
**5696**

### Re: Problem 264

Yes, Rodinio, I believe your result for 5000 is correct; but I cannot offer any advice whether you should or should not port it to C