## Search found 18 matches

- Tue Mar 06, 2018 4:52 pm
- Forum: Combinatorics
- Topic: BBC Article - Lego arrangements
- Replies:
**1** - Views:
**991**

### BBC Article - Lego arrangements

I read this morning a story about Lego on the BBC news app. It contained the following nugget: 915,103,765 - the number of ways to combine six two-by-four Lego bricks of the same color That seemed surprisingly high a number to me. Assume the following: the bricks are indistinguishable (all the same ...

- Fri Nov 17, 2017 1:59 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 613
- Replies:
**7** - Views:
**2023**

### Re: Problem 613

Thanks v6ph1 and LilStlker,

Finally got back to the problem and knocked it out in 10 minutes or so!

Finally got back to the problem and knocked it out in 10 minutes or so!

- Sun Nov 05, 2017 5:42 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 613
- Replies:
**7** - Views:
**2023**

### Problem 613

Not sure this matters, since it only applies to two points on the perimeter of the triangle, but what happens if the ant lands on the intersection of the hypotenuse and either of the short sides? There is a range of directions that involve the ant exiting the triangle crossing both the hypotenuse an...

- Thu Sep 14, 2017 1:38 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 146
- Replies:
**18** - Views:
**5478**

### Re: Problem 146

hk,

Thanks for your reply - only just saw it!

Thanks for your reply - only just saw it!

- Sun Aug 27, 2017 5:39 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 146
- Replies:
**18** - Views:
**5478**

### Re: Problem 146

Curious about this and similar problems: Regarding the one-minute rule, do you generate the primes you need in your code and thus include the time to generate the primes in the one-minute, or do you use a pre-populated list of primes? It takes my machine and R and what I think is a fairly good imple...

- Mon Jun 26, 2017 12:10 am
- Forum: News, Suggestions, and FAQ
- Topic: Suggestion: Searching Problems
- Replies:
**4** - Views:
**978**

### Re: Suggestion: Searching Problems

Jaap and hk,

Thanks for responding. Since I always log in it never occurred to me that Google would be able to crawl all the problems anyway.

I guess I should say "doh!".

Thanks for responding. Since I always log in it never occurred to me that Google would be able to crawl all the problems anyway.

I guess I should say "doh!".

- Sat Jun 24, 2017 6:22 pm
- Forum: News, Suggestions, and FAQ
- Topic: Suggestion: Searching Problems
- Replies:
**4** - Views:
**978**

### Suggestion: Searching Problems

I hope this hasn't been asked before - I searched the forum a couple different ways and didn't find the topic. Is it possible to implement some kind of search functionality in the archived problems? For example, one might search for "collatz" or "nim" to find related problems. I know this would help...

- Mon Jun 19, 2017 9:00 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 606
- Replies:
**8** - Views:
**1615**

### Re: Problem 606

Thanks MuthuVeerappanR, V6ph1, and yourmaths. I appreciate your feedback! My guess is I need to better understand how to take advantage of the modular arithmetic part of problems such as this one. "Last 9 digits ..." essentially means answer mod 10^9. So I think I need try to do the S(10^12) example...

- Fri Jun 16, 2017 4:23 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 606
- Replies:
**8** - Views:
**1615**

### Problem 606

I know this is quite soon after the problem was posted, but I am finding myself in recent problems coming up against issues to do with the sheer size of the numbers involved. Trying very hard not to give anything away, here's what I mean regarding this problem: I have found an approach that gives me...

- Sun Feb 15, 2015 2:33 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 500
- Replies:
**2** - Views:
**1018**

### Re: Problem 500

Message received - Thanks!

- Sat Feb 07, 2015 7:49 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 500
- Replies:
**2** - Views:
**1018**

### Problem 500

Can someone confirm that the answer for 2^10,000 divisors, mod 500,500,507 is <removed by moderator>?

(Link to problem added by moderator:

(Link to problem added by moderator:

**Problem 500**(View Problem))- Wed Jan 15, 2014 5:27 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 445
- Replies:
**10** - Views:
**2896**

### Re: Problem 445

Last step, some more interesting retractions: {n=6, a=3, b=0} is a retraction because f(0) = 0 => f(f(0)) = f(0) = 0 and f(2) = f(4) = 0 f(3) = 3 => f(f(3)) = f(3) = 3 and f(1) = f(5) = 3 {n=6, a=4, b=3} is a retraction because f(0) = 0 => f(f(0)) = f(0) = 0 and f(3) = 0 f(2) = 2 => f(f(2)) = f(2) =...

- Tue Jan 14, 2014 3:35 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 445
- Replies:
**10** - Views:
**2896**

### Re: Problem 445

Thank you mdean. Next step (I assume the following is so trivial as not to be a spoiler) ... a=1 and b=0 => f(x) = x mod n x <= n, so f(x) = x when 0 <= x < n and f(x) = 0 when x =n Therefore, f(f(x)) = f(x) for all integer values of x, 0 <= x <= n. i.e. f(x) is a retraction for all sets {n, a=1, b=...

- Mon Jan 13, 2014 2:06 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 445
- Replies:
**10** - Views:
**2896**

### Problem 445

I think I just don't get modular arithmetic! I would like to get a better understanding without spoiling this problem so I will try baby steps. Can someone verify the following chain of logic: Consider n=12, a=10, b=3 and x=6 f(x) = (10.6 + 3) mod 12 = 63 mod 12 = 3 f(f(x)) = f(3) = (10.3 + 3) mod 1...

- Wed Apr 04, 2012 6:18 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 054
- Replies:
**63** - Views:
**16530**

### Re: Problem 054

EDIT: Figured it out, problem with 3 of a kind. PLUS: Suggestion to anyone else trying to verify their ranking algo: see if your counts of each type of hand (2 pair, 3 of a kind, etc) matches expectations - I found I had a count of 3 of a kind that had only a 0.08% probability. You can find the prob...

- Thu Feb 02, 2012 3:55 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 369
- Replies:
**12** - Views:
**3301**

### Re: Problem 369

Thank you TripleM, your hint was understood - I see the source of my double-counting! Of course, that doesn't mean I can solve the problem! May I suggest if anyone else is having the same problem as me, simplify the deck to ranks 1-3 and suits H and S, and define a badugi' as 2 cards of different ra...

- Tue Jan 31, 2012 11:06 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 369
- Replies:
**12** - Views:
**3301**

### Re: Problem 369

Still not getting it ... so two questions 1) 5 cards, all different ranks, obviously 2 of same suit, e.g 2H, 6H, 3S, 4D, 5C. Does this only count as 1 badugi-containing hand, even though I can form 2 different badugis: 2H, 3S, 4D, 5C AND 6H, 3S, 4D, 5C? 2) 5 cards containing 1 badugi, can the fifth ...

- Tue Jan 31, 2012 6:32 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 369
- Replies:
**12** - Views:
**3301**

### Problem 369

Having trouble making sure I understand a Badugi, because I can't match the 5 card example. My understanding: Regardless the number of cards in the hand (from 4 to 13), I must be able to form at least one set of 4 cards with the following property: Each card in the set of 4 has a different rank and ...