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Project Euler Solutions

Project Euler 207 Solution

Project Euler 207 Solution

Project Euler 207: Integer partition equations


Problem Description

For some positive integers k, there exists an integer partition of the form   4t = 2t + k,
where 4t, 2t, and k are all positive integers and t is a real number.

The first two such partitions are 41 = 21 + 2 and 41.5849625… = 21.5849625… + 6.

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 km.
Thus P(6) = 1/2.

In the following table are listed some values of P(m)

   P(5) = 1/1
   P(10) = 1/2
   P(15) = 2/3
   P(20) = 1/2
   P(25) = 1/2
   P(30) = 2/5
   …
   P(180) = 1/4
   P(185) = 3/13

Find the smallest m for which P(m) < 1/12345

Project Euler 207 Solution

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Answer

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Afterthoughts

    No afterthoughts yet.

Partition is perfect when first element is a power of 2.
I think the original problem statement is intentionally obtuse to make the problem appear more difficult.,
For L = 123456; Smallest m for which P(m) < 1 / 123456 = 6721458093506 [/comments]
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