## Search found 13 matches

- Wed Feb 13, 2019 7:24 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 250
- Replies:
**14** - Views:
**2970**

### Re: Problem 250

Suppose I have a set that totals to 250, say {1^1 + 249^249}. Would there exist a similar set with a zero, say {1^1 + 249^249 + 250^250}? In fact, since there are 25,025 ways where x^x % 250 = 0, then for any set that total to 250 that doesn't have any zeros, there exist many other sets that use the...

- Tue Feb 12, 2019 9:02 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 223 & 224
- Replies:
**22** - Views:
**7302**

- Tue Feb 12, 2019 6:01 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 223 & 224
- Replies:
**22** - Views:
**7302**

### Re: Problem 223 & 224

Can somebody please confirm the number of solutions for perimeter < 10^4 is equal to 13656 For some reason, I can't write even a brute force algorithm that gets even close to 13,656 for 10^4. I'm checking every variation of a,b,c where a <= b <= c, c < a + b, and a + b + c <= 10^4, but I'm only get...

- Thu Aug 30, 2018 6:41 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 504
- Replies:
**12** - Views:
**3152**

### Re: Problem 504

Just to confirm, if a=b=c=d=2, then because there are four points that lie exactly on the diagonal edges of the quadrilateral, they would not be counted and so this would strictly contain only 5 lattice points?

- Tue Aug 21, 2018 5:17 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 113
- Replies:
**1** - Views:
**329**

### Problem 113

To confirm, the following numbers are all

1,2,3,4,5,6,7,8,9,11,22,33,44,55,66,77,88,99,111,222,333,444,555,666,777,888,999, etc

*both*increasing and decreasing and therefore are*not*bouncy, correct?1,2,3,4,5,6,7,8,9,11,22,33,44,55,66,77,88,99,111,222,333,444,555,666,777,888,999, etc

- Thu Aug 16, 2018 9:12 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 185
- Replies:
**21** - Views:
**6394**

### Re: Problem 185

Gene, if you're still looking at this problem 6 years later :lol:, I think the reason is that if there is just one correct digit, then there are only 16 possibilities. However, if there are three correct digits, then there are 560 possibilities. So by starting your tree at the top with the minimal a...

- Wed Aug 01, 2018 5:10 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 381
- Replies:
**9** - Views:
**3642**

### Re: Problem 381

Thank you! That was what I needed. I was multiplying too high and went above a 64-bit number. Fixed and solved

- Tue Jul 31, 2018 11:44 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 381
- Replies:
**9** - Views:
**3642**

### Re: Problem 381

I'm getting an answer, but it's not correct. However I'm able to produce the correct number when p < 10^2. Can someone confirm if the next few powers of 10 are correct?

p<10^3 =

p<10^4 = [removed by moderator]

p<10^5 =

p<10^3 =

p<10^4 = [removed by moderator]

p<10^5 =

- Fri Aug 26, 2016 6:52 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 094
- Replies:
**47** - Views:
**13855**

### Re: Problem 094

Got the help I needed. Thanks

- Tue Jul 19, 2016 10:49 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 301
- Replies:
**9** - Views:
**1502**

### Re: Problem 301

Realized I was checking for wins and not losses. Either way though, I had to change it completely. I have it now, thank you!v6ph1 wrote:They are all wrong.

n<=2^0: 1

n<=2^1: 2

- Tue Jul 19, 2016 9:32 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 301
- Replies:
**9** - Views:
**1502**

### Problem 301

I have an algorithm that gets me an answer that is not correct, but I feel like it should be. Can someone confirm if they get the same results where n <= 2^x?

Code: Select all

```
f(1) = 1
f(2) = 1
f(3) = 3
f(4) = 8
f(5) = 18
f(6) = 39
f(7) = 81
f(8) = 166
f(9) = 336
f(10) = 677
```

- Tue Sep 29, 2015 11:57 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 057
- Replies:
**2** - Views:
**769**

### Re: Problem 057

Ah, didn't even realize that. I bet the numbers are too large and resetting back to 0. Thanks!

- Tue Sep 29, 2015 11:16 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 057
- Replies:
**2** - Views:
**769**

### Problem 057

I'm getting an answer that isn't the right one, and I have no idea why. Can someone confirm the following? Expansion - Fraction #9 = 3363/2378 #10 = 8119/5741 #11 = 19601/13860 ... #500 = 5297586898435161161/8492144183025380461 #501 = 3835131190776370467/13789731081460541622 #502 = 12967849279987902...