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Let $n>0$ be an integer. Consider all partitions of $n$, i.e. all possible ways of writing $n$ as a finite sum of positive integers, $$n=n_1+n_2+\cdots+n_k.$$ What partition maximizes the product $n_1n_2\cdots n_k$?

Examples:

  • $2=2$
  • $3=3$
  • $4=2+2$
  • $5=3+2$
  • $6=3+3$
  • $7=3+2+2$
  • $8=3+3+2$
  • $9=3+3+3$
  • $10=3+2+2$
  • $20=3+3+3+3+3+3+2$
  • $30=3+3+3+3+3+3+3+3+3+3$
  • $40=3+3+3+3+3+3+3+3+3+3+3+3+2+2$
  • $50=3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+2$

Conjecture: It is the partition that has the maximum number of $3$'s among those consisting only of $3$'s and $2$'s.

Simon Parker
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1 Answers1

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Here is a rough idea. The problem can be divided into two parts. Firstly, given a $n$, for a fixed number of partitions $k$, it is not hard to find that all of the partitions should be equal. However, since we are dealing with integers, we might need to find the nearest integers.

Having the first part, we would like to know for which $k$ we get the optimum solution. we want to optimize the logarithm of the product instead. Please note that I have simplified things by letting numbers to be real.

$ln(n)=\sum_{i=1}^{k}ln(x_i)$

considering all $x_i$ are the same, we get

$ln(n)=k*ln(\frac{n}{k})$

taking the derivative of the function with respect to $k$ and putting it equal to zero, we get

$ln(\frac{n}{k})=1$

Therefore

$k\simeq\frac{n}{e}$

finally

$\frac{n}{k}=e\simeq2.7$

Med
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