## Problem 207

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kkwweett
Posts: 2
Joined: Tue Nov 04, 2008 1:42 am

### Problem 207

Hi everyone,

"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

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

Thank you and sorry for my distraction !

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

### Problem 207

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?

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

### Re: Problem 207

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

### 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?
49.157.5694.1125

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

### Re: Problem 207

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

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?

jaap
Posts: 550
Joined: Tue Mar 25, 2008 3:57 pm
Contact:

### Re: Problem 207

...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 $P(6)=\frac{1}{2}$.