Project Euler Forum

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

4^(1.20337385442234...) and 2^(1.20337385442234...) are not a positive integers.

Thank you and sorry for my distraction !

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?

well, I found viewtopic.php?f=50&t=1189&hilit=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?

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

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?

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