## Search found 165 matches

- Wed Dec 18, 2019 11:27 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 678
- Replies:
**0** - Views:
**1116**

### Problem 678

Does x = c1^f1 = c2^f2 count as one case or two?

- Tue Feb 12, 2019 7:00 pm
- Forum: Number Theory
- Topic: Project Euler Factoring Challenge
- Replies:
**4** - Views:
**8065**

### Re: Project Euler Factoring Challenge

Haven't tried it, but can surely recommend the book "The Joy Of Factoring" (Samuel Wagstaff Jr.)

- Tue Feb 12, 2019 5:22 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 263
- Replies:
**27** - Views:
**8644**

### Re: Problem 263

Can I safely assume there are 28,388 sexy triple-pairs (n-9,n-3)-(n-3,n+3)-(n+3,n+9) below 10

^{9}?- Thu Jul 19, 2018 10:32 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 308
- Replies:
**9** - Views:
**3651**

### Re: Problem 308

Apparently it's the slow branch then. Thanks

- Thu Jul 19, 2018 6:39 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 308
- Replies:
**9** - Views:
**3651**

### Re: Problem 308

Is that vote because you've already solved this? Or just a guess...

- Thu Jul 19, 2018 2:27 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 308
- Replies:
**9** - Views:
**3651**

### Re: Problem 308

There seem to exist two valid versions. Version A takes 707 steps to find 2^7, Version B takes 710 steps.

On which of those does the correct solution rely?

On which of those does the correct solution rely?

- Thu Apr 12, 2018 8:05 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 622
- Replies:
**6** - Views:
**4560**

### Re: Problem 622

Thanks

- Thu Apr 12, 2018 7:54 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 622
- Replies:
**6** - Views:
**4560**

### Problem 622

From what I gather from the problem description, the outcome of s(n) is 0 if n is odd, else some value > 0. Is this right?

- Thu Apr 05, 2018 5:47 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 141
- Replies:
**27** - Views:
**11161**

### Re: Problem 141

My thoughts concerning the sample numbers from the description: 9 is in the list because with d=2 we get q=4, r=1. For this [1,2,4] there's the factor 2 forming this geometric sequence.

Am I right with this? Or am I completely going wrong?

Am I right with this? Or am I completely going wrong?

- Thu Nov 09, 2017 12:22 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 207
- Replies:
**6** - Views:
**2866**

### Re: Problem 207

Those numbers should be correct but the algo wouldn't work for some special cases. Changed it and solved it

- Thu Nov 09, 2017 12:06 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 207
- Replies:
**6** - Views:
**2866**

### Re: Problem 207

Hello there,

I've run some tests with my algo which seems to work, but I'm not quite sure if I got the description right. My tests suggest the smallest m for which (A) P(m) < 1/75 is 4xxx52, (B) P(m) < 1/100 is 8xxx02, (C) P(m) < 532 is 4xxxx610.

Are these numbers even close to correct?

I've run some tests with my algo which seems to work, but I'm not quite sure if I got the description right. My tests suggest the smallest m for which (A) P(m) < 1/75 is 4xxx52, (B) P(m) < 1/100 is 8xxx02, (C) P(m) < 532 is 4xxxx610.

Are these numbers even close to correct?

- Tue Jan 10, 2017 8:23 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 581
- Replies:
**8** - Views:
**1708**

### Problem 581

The problem and its description are pretty much straight forward. A brute-force algorithm may work for p-smooth numbers with small p. I've tried something different and tried several sets of primes [2, 3...]. Following the description of this problem, might I ask if I'm right with the following: the...

- Tue Dec 06, 2016 7:49 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 365
- Replies:
**9** - Views:
**2071**

### Re: Problem 365

It is

- Thu Nov 24, 2016 8:15 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 365
- Replies:
**9** - Views:
**2071**

### Re: Problem 365

I suspect there are 20[...]50 triplets to process. Is that somewhat close?

- Mon Nov 07, 2016 4:27 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 571
- Replies:
**10** - Views:
**2667**

### Re: Problem 571

Thanks @DJohn for pointing this out. Solved it

- Mon Oct 17, 2016 12:23 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 571
- Replies:
**10** - Views:
**2667**

### Re: Problem 571

So it's the sum of the ten smallest pandigital numbers in base n, and those numbers have to be pandigital in all bases < n as well.

- Fri Oct 14, 2016 7:15 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 571
- Replies:
**10** - Views:
**2667**

### Problem 571

I've read and re-read the description of no. 571 again and again. Also I've made up a little programme which helped me find the smallest n-super-pandigital numbers for 2 <= n <= 10. I also get the results proposed in the description, but the sum of them [2..10] is something completely different. In ...

- Sat Apr 02, 2016 8:57 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 457
- Replies:
**2** - Views:
**2263**

### Re: Problem 457

With primes p < 1000, is it correct to have 80 primes giving R(p) != 0

[edit]

Yes it is...

[edit]

Yes it is...

- Fri Dec 04, 2015 10:11 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 231
- Replies:
**1** - Views:
**1575**

### Re: Problem 231

It wouldn't...

- Thu Dec 03, 2015 11:29 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 231
- Replies:
**1** - Views:
**1575**

### Problem 231

In the process of eliminating shitty flaws

Would the sum of prime factors of

Would the sum of prime factors of

^{54321}C_{12345}be = 54,877,718?