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Consider the number of ways to ordered partition an integer $x > 4$ into $x-4$ ones and $2$ twos. In each of these partitions, we want to split them into at least two groups such that the sum of each group is the same. For example, for $x=8$, we can split the ordered partition $\lbrace 2,2,1,1,1,1 \rbrace$ into $\lbrace (2,2),(1,1,1,1) \rbrace$ as both groups have the same sum.

We want to find all $x$ such that it is possible to do this for all possible ordered partitions of $x$ in this manner.

Through brute force, I found the first example that works is $x=6$. Clearly $x$ cannot be prime. I suspect that it is only possible for $x$ that can only be prime factorized as the product of distinct primes, such as $6 = 2 \times 3, 30 = 2 \times 3 \times 5, \ldots$ as these are immune to a two "blocking" a factor. This is why $x=8$ does not work, since the arrangement $\lbrace 2,1,2,1,1,1 \rbrace$ prevents a factor of $4$ or $2$ from occurring.

Any suggestions or thoughts are appreciated. Perhaps as a prime number gets big enough, the number of $2$'s will not matter. It may also not work for prime powers as well.

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  • $\begingroup$ Is it necessary that there are two groups, or can there be $3$ groups, each adding up to $x//3?$ $\endgroup$ Commented Oct 23 at 17:11
  • $\begingroup$ @ThomasAndrews "at least 2" will include the possibility of 3 groups. $\endgroup$ Commented Oct 23 at 17:40
  • $\begingroup$ if there were only two, that would trivialize the problem, wouldn't it $\endgroup$ Commented Oct 23 at 17:54
  • $\begingroup$ USAMTS Year 37 Round 2 $\endgroup$ Commented Nov 3 at 15:24

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