the problem statement saying that :
"The sum of the primes q <100 such that P(q)=1 equals 233."
however my calculations saying that the sum is 630 :
Expand
Problem 333
Forum rules
As your posts will be visible to the general public you are requested to be thoughtful in not posting anything that might explicitly give away how to solve a particular problem.
This forum is NOT meant to discuss solution methods for a problem.
See also the topics:
Don't post any spoilers
Comments, questions and clarifications about PE problems.
As your posts will be visible to the general public you are requested to be thoughtful in not posting anything that might explicitly give away how to solve a particular problem.
This forum is NOT meant to discuss solution methods for a problem.
In particular don't post any code fragments or results.
Don't start begging others to give partial answers to problems
Don't ask for hints how to solve a problem
Don't start a new topic for a problem if there already exists one
Don't start begging others to give partial answers to problems
Don't ask for hints how to solve a problem
Don't start a new topic for a problem if there already exists one
See also the topics:
Don't post any spoilers
Comments, questions and clarifications about PE problems.
-
elr
- Posts: 67
- Joined: Thu Apr 09, 2009 9:47 am
Problem 333
i think that i miss something
the problem statement saying that :
"The sum of the primes q <100 such that P(q)=1 equals 233."
however my calculations saying that the sum is 630 :
the problem statement saying that :
"The sum of the primes q <100 such that P(q)=1 equals 233."
however my calculations saying that the sum is 630 :
i am wondering how a sum of 233 could be made ?
-
jaap
- Posts: 588
- Joined: Tue Mar 25, 2008 3:57 pm
- Contact:
Re: Problem 333
You have some false positives, for example:
31 = 27 + 4 = 9 + 6 + 16
41 = 9 + 32 = 27 + 6 + 8
and some missing ones, for example:
23 = 9 + 6 + 8
It looks like you are only using powers of 2 and powers of 3, and missing out terms like 6 which are neither.
31 = 27 + 4 = 9 + 6 + 16
41 = 9 + 32 = 27 + 6 + 8
and some missing ones, for example:
23 = 9 + 6 + 8
It looks like you are only using powers of 2 and powers of 3, and missing out terms like 6 which are neither.
-
gsidebottom
- Posts: 1
- Joined: Tue May 03, 2011 4:40 am
Re: Problem 333
I'm having problems with this one as well. In particular, my program is getting the sum of the primes q <100 such that P(q)=1 equals 490. Below are the primes less than 100 with unique partitions as far as I understand. Anybody see what my problem is?
Expand
-
jaap
- Posts: 588
- Joined: Tue Mar 25, 2008 3:57 pm
- Contact:
Re: Problem 333
71 = 27 + 36 + 8gsidebottom wrote:71=32+27+12
89=81+8
97=81+16
89 = 27 + 18 + 12 + 32
97 = 27 + 6 + 64
-
PhilLeTaxi
- Posts: 20
- Joined: Wed Mar 05, 2014 12:54 pm
- Location: Paris, France
-
vamsikal3
- Posts: 89
- Joined: Sat Oct 01, 2016 9:25 am
Re: Problem 333
<deleted post>
Last edited by vamsikal3 on Fri Nov 27, 2020 3:11 am, edited 2 times in total.
my friend key --> 990813_OZPwQtCjkD6KlvxirOoTSZxccMFsuw1L
-
hk
- Administrator
- Posts: 12831
- Joined: Sun Mar 26, 2006 10:34 am
- Location: Haren, Netherlands
-
vamsikal3
- Posts: 89
- Joined: Sat Oct 01, 2016 9:25 am
Re: Problem 333
<deleted post>
Last edited by vamsikal3 on Fri Nov 27, 2020 3:11 am, edited 2 times in total.
my friend key --> 990813_OZPwQtCjkD6KlvxirOoTSZxccMFsuw1L
-
Jochen_P
- Posts: 55
- Joined: Mon Oct 05, 2009 10:47 am
- Location: Stuttgart, Germany
Re: Problem 333
Resuming the problems after a long break.
My terms for q <100 are
[2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 81, 96]
my result for P(q)=1 for q<100 is 228 (instead of 233)
Has anybody any clue where I'm going wrong?
My primes with P(q) = 1 are
5
7
13
17
23
43
59
61
My terms for q <100 are
[2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 81, 96]
my result for P(q)=1 for q<100 is 228 (instead of 233)
Has anybody any clue where I'm going wrong?
My primes with P(q) = 1 are
5
7
13
17
23
43
59
61
-
traxex
- Posts: 66
- Joined: Thu Oct 19, 2017 1:30 pm
Re: Problem 333
Perhaps you are missing two small primes that sum to 5?
Technically, everyone is full of himself.
-
Jochen_P
- Posts: 55
- Joined: Mon Oct 05, 2009 10:47 am
- Location: Stuttgart, Germany
Re: Problem 333
Aye, thanx a lot!
I kind of left them out by purpose.. thought once again incorrectly that a partition must consist of n>1 elements. m(
Now to make this bugger efficient. Runs way too long to get P(q)=1 for q<10**6
I kind of left them out by purpose.. thought once again incorrectly that a partition must consist of n>1 elements. m(
Now to make this bugger efficient. Runs way too long to get P(q)=1 for q<10**6
-
Jochen_P
- Posts: 55
- Joined: Mon Oct 05, 2009 10:47 am
- Location: Stuttgart, Germany
Re: Problem 333
... way too long to get even sum P(q)=1 for q<10**3.
4600 in a bit more than an hour m(
Can't think of any other good way to optimize that anymore, there are way too many partitions to check.
There has to be a simple rule/formula for excluding terms in this, but damn, do I suck at maths
4600 in a bit more than an hour m(
Can't think of any other good way to optimize that anymore, there are way too many partitions to check.
There has to be a simple rule/formula for excluding terms in this, but damn, do I suck at maths
-
Col_Kernel
- Posts: 1
- Joined: Sun Oct 15, 2017 11:17 am
Re: Problem 333
Is it possible that "Confimed." above refers to something in the deleted post? First I believed it refers to the preceding question for q < 10000 and started to debug my algorithm. But I get 24853 for q < 10000 (and my answer for 10^6 got accepted ).
).