Problem 207

A place to air possible concerns or difficulties in understanding ProjectEuler problems. This forum is not meant to publish solutions. This forum is NOT meant to discuss solution methods or giving hints how a problem can be solved.
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.

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


See also the topics:
Don't post any spoilers
Comments, questions and clarifications about PE problems.
Post Reply
kkwweett
Posts: 2
Joined: Tue Nov 04, 2008 1:42 am

Problem 207

Post by kkwweett »

Hi everyone,

please help me with this sentence :

"The first two such partitions are 4^1 = 2^1 + 2 and 4^(1.5849625...) = 2^(1.5849625...) + 6."

Why 4^1 = 2^1 + 2 and 4^(1.20337385442234...) = 2^(1.20337385442234...) + 3 aren't the first two such partitions ?

Thanks
Ikcelaks
Posts: 28
Joined: Wed Oct 15, 2008 9:08 pm

Re: Problem 207 Clarification

Post by Ikcelaks »

4^(1.20337385442234...) and 2^(1.20337385442234...) are not a positive integers.
kkwweett
Posts: 2
Joined: Tue Nov 04, 2008 1:42 am

Re: Problem 207 Clarification

Post by kkwweett »

Thank you and sorry for my distraction !
odr
Posts: 2
Joined: Sun Sep 26, 2010 7:14 pm

Problem 207

Post by odr »

In problem Problem 207 (View Problem) I don't understand why the second k = 6?

What's wrong with k=3, t=1.2033738544223431; k=4, t=1.3570186368605763 and so on?


Image
odr
Posts: 2
Joined: Sun Sep 26, 2010 7:14 pm

Re: Problem 207

Post by odr »

User avatar
Oliver1978
Posts: 166
Joined: Sat Nov 22, 2014 9:13 pm
Location: Erfurt, Germany

Re: Problem 207

Post by Oliver1978 »

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?
49.157.5694.1125
User avatar
Oliver1978
Posts: 166
Joined: Sat Nov 22, 2014 9:13 pm
Location: Erfurt, Germany

Re: Problem 207

Post by Oliver1978 »

Those numbers should be correct but the algo wouldn't work for some special cases. Changed it and solved it :)
49.157.5694.1125
matthew.townsend
Posts: 1
Joined: Mon Feb 24, 2020 10:27 pm

Re: Problem 207

Post by matthew.townsend »

Partitions where t is also an integer are called perfect.
For any m ≥ 1 let P(m) be the proportion of such partitions that are perfect with k ≤ m.
Thus P(6) = 1/2.
I don't get how P(6) can be 1/2.

k must be positive, which excludes non-positive values for t.

k = 1 has no solution where t is also an integer
k = 2 has a solution with t = 1
k = 3 has no solution where t is also an integer
k = 4 has no solution where t is also an integer
k = 5 has no solution where t is also an integer
k = 6 has no solution where t is also an integer

The example given
4^1.5849625... = 2^1.5849625... + 6
evaluates to 9 = 3 + 6, which looks like a hint, but the problem says a partition is perfect where t is also an integer, which it clearly isn't. 4^t and 2^t are though.

Does this problem mean to ask for cases where 4^t and 2^t are each themselves positive integers, regardless of whether or not t specifically is an integer?
User avatar
jaap
Posts: 553
Joined: Tue Mar 25, 2008 3:57 pm
Contact:

Re: Problem 207

Post by jaap »

...where $4^t$, $2^t$, and $k$ are all positive integers and $t$ is a real number.
For any m ≥ 1 let P(m) be the proportion of such partitions that are perfect with k ≤ m.
So there are two partitions, $4^1 = 2^1 + 2$ and $4^{1.5849625...} = 2^{1.5849625...} + 6$ with $k\le 6$, only one of which has an integer value for $t$. So half the solutions so far are perfect. Hence $P(6)=\frac{1}{2}$.
Post Reply