Yep, the point (x,y,z) must be on the surface of the sphere and its coordinates (i.e. x, y and z) must be integers.LarryBlake wrote:Is that correct?

## Search found 261 matches

- Sun Apr 10, 2011 2:16 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 332
- Replies:
**4** - Views:
**3040**

### Re: Problem 332

- Sun Mar 20, 2011 7:53 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 329
- Replies:
**18** - Views:
**5651**

### Re: Problem 329

Yes, there is a croak for the first square too (the wording has not been changed).

Thanks for pointing out the spelling error - now fixed.

Thanks for pointing out the spelling error - now fixed.

- Wed Mar 16, 2011 9:08 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 328
- Replies:
**27** - Views:
**8489**

### Re: Problem 328

@

A couple of days ago, you asked "if the solutions so far are meeting the 1 minute rule".

How do you feel now that your own solution is "under 2 sec" ?

**sivakd**: But, as I can see, now you've managed to solve it (and make the fastest-20 list too !). Well doneA couple of days ago, you asked "if the solutions so far are meeting the 1 minute rule".

How do you feel now that your own solution is "under 2 sec" ?

- Mon Mar 14, 2011 2:45 pm
- Forum: Programming languages
- Topic: Fortran
- Replies:
**2** - Views:
**2470**

### Re: Fortran

You can PM your Fortran code to me, if you wish. I very rarely use Fortran nowadays, but my experience with it goes back several decades

- Mon Mar 14, 2011 7:17 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 328
- Replies:
**27** - Views:
**8489**

### Re: Problem 328

Well said TripleM. No further intermediate results should be given (at this stage, anyway). Sorry, kingvash!

Let's not forget that the idea behind this forum is to help with

Let's not forget that the idea behind this forum is to help with

*understanding*the problems (though I plead guilty to occasionally going beyond that).- Sun Mar 13, 2011 10:11 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 328
- Replies:
**27** - Views:
**8489**

### Re: Problem 328

The result posted by sivakd for the sum 1 to 200, seems correct.

However, the result posted by Cifko for the sum 1 to 10000 looks too high.

However, the result posted by Cifko for the sum 1 to 10000 looks too high.

- Sun Mar 13, 2011 2:32 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 328
- Replies:
**27** - Views:
**8489**

### Re: Problem 328

Yes. For n=8, with "4" as the first question, the worst-case has a cost of 16 (questions: "4", "5" and "7"). The "binary search" strategy (first question "4", second question "2" or "6" depending on the first answer etc) has a worst-case cost of 17 as described in 328 . Neither of those is as good a...

- Thu Feb 24, 2011 5:54 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 325
- Replies:
**7** - Views:
**3483**

### Re: Problem 325

Obviously, it's a winning configuration for the first player.

(S)he can win by taking 2*3 = 6 stones from the larger pile. That will leave 3,4 for the other player, which, as stated in the

(S)he can win by taking 2*3 = 6 stones from the larger pile. That will leave 3,4 for the other player, which, as stated in the

**Problem 325**(View Problem) is a losing configuration for the player whose turn is to play.- Sat Feb 19, 2011 3:21 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 260
- Replies:
**9** - Views:
**4356**

### Re: Problem 260

i is the index of the loosing configuration(s). E.g. if (0,1,2) and (1,3,3) are the only losing configurations you need to consider, then you can assing the index i=1 to (0,1,2) and the index i=2 to (1,3,3). (The actual order is of course unimportant). Then, x 1 =0, y 1 =1, z 1 =2, x 2 =1, y 2 =3, ...

- Thu Feb 17, 2011 4:42 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 048
- Replies:
**15** - Views:
**5161**

### Re: Problem 048

A. Please don't start a new thread here, if there is already one for the specific problem.

B. W.r.t. efficient methods : I suggest you read through the problem's forum (in the main site), where those who have solved it before you, are discussing their methods.

B. W.r.t. efficient methods : I suggest you read through the problem's forum (in the main site), where those who have solved it before you, are discussing their methods.

- Wed Feb 16, 2011 7:31 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 005
- Replies:
**49** - Views:
**14809**

### Re: Problem 005

You can send a PM to me with details of what exactly it is that you don't understand in the pdf.Inglonias wrote:.... I looked at the overview, but I didn't understand it.

Then, I might be able to explain it better.

- Sun Jan 09, 2011 10:21 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 319
- Replies:
**23** - Views:
**8098**

### Re: Problem 319

No, but you are close. Here is the complete list for n=3:
You can't ask for more, can you?

Good luck!

Code: Select all

```
2 4 8
2 4 9
2 4 10
2 4 11
2 5 11
2 5 12
2 5 13
2 5 14
2 6 14
2 6 15
2 6 16
2 6 17
2 6 18
2 7 18
2 7 19
2 7 20
2 7 21
2 7 22
2 8 22
2 8 23
2 8 24
2 8 25
2 8 26
```

Good luck!

- Sat Jan 08, 2011 11:08 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 319
- Replies:
**23** - Views:
**8098**

### Re: Problem 319

It does, if you look carefully atstarblue wrote:I don't see how {2, 6} satisfies the third condition:

$$2^6=64 \not< 49 = (6+1)^2$$

**Problem 319**(View Problem): the exponents are i and j,

*not*x

_{i}and x

_{j}.

So, 2

^{2}<(6+1)

^{1}and 6

^{1}<(2+1)

^{2}, both of which are true.

- Fri Dec 31, 2010 10:35 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 316
- Replies:
**32** - Views:
**14332**

### Re: Problem 316

Surprisingly this problem has low private-forum-posts/solvers ratio. Perhaps due to fewer ways to solve it? I tend to disagree : Right now, #316 has 97 correct answers and 25 posts, #314 has 95 correct solutions and 22 posts, #311 has 108 correct solutions and 35 posts, #312 has 182 correct solutio...

- Mon Dec 27, 2010 6:41 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 316
- Replies:
**32** - Views:
**14332**

### Re: Problem 316

@texane : There seem to be at least two errors in your code: (a) When state==1 && n==5, the inner loop returns state=0. Imo, it should return state=1. (b) When 535 is eventually found and you break out of the inner loop the variable "count" is not the index of the first digit, as the problem states...

- Fri Dec 03, 2010 4:25 pm
- Forum: Clarifications on Project Euler Problems
- Topic: problem 307
- Replies:
**16** - Views:
**5675**

### Re: problem 307

@mukeshtiwari : Yes, there is something wrong with your logic: as stijn263 said, the probabilities of the arrangements you have listed are not equal. Consider the following: If you throw two normal dice, there are 11 possible values for their sum: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Does that mea...

- Thu Dec 02, 2010 5:56 pm
- Forum: News, Suggestions, and FAQ
- Topic: Bug in cache refresh for the problems page?
- Replies:
**2** - Views:
**1466**

### Re: Bug in cache refresh for the problems page?

It's ok now...

- Wed Dec 01, 2010 10:21 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 189
- Replies:
**9** - Views:
**3994**

### Re: Problem 189

Yes, your answer for 16 triangles, seems to be correct

- Thu Nov 25, 2010 8:00 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 003
- Replies:
**128** - Views:
**42558**

### Re: Problem 003

I suggest you try the Sieve of Eratosthenes and some of its variants / improvements. You'll suprised !

- Mon Nov 15, 2010 7:16 am
- Forum: Programming languages
- Topic: c++ numbers out of range
- Replies:
**9** - Views:
**6438**

### Re: c++ numbers out of range

Welcome to PE latinpower! C++ also has 64-bit integers ("long long" or "__int64" depending on which implementation you use). As a principle, you don't need any bigInt or bigNum libraries to solve the PE problems (but you do need to know the limitations imposed by each type and take those limitations...