Partition integer range (1 to n) as multiples of divisors of n

Question

Positive integers everywhere.

I need to partition a set {1, 2, 3, ..., n} into disjoint subsets, one for each (not necessarily prime) divisor of n, such that

any divisor d divides every element in its allocated subset
no other divisor (> d) of n divides any element in that subset.

I only need the sizes of the subsets, not the subsets themselves.

It is kind of confusing to comprehend. Can you help me out?

Note - A friend showed me this problem from their Mathematics paper. I'm not sure where it came from. So if you can find a source, can you put it here?

lhf · Accepted Answer · 2021-05-16T14:22:52.850

1

Hint: Let $X_d = \{ x \in \{1,2,\dots,n\} : ord_n(x)=n/d \}$.

Here, $ord_n(x)$ is the additive order of $x$ mod $n$.

edited May 16 '21 at 14:22

answered May 16 '21 at 14:04

lhf

221,500

I thought point #2 made this useless. – nomad May 16 '21 at 14:13
@nomad, no, because $d \in X_d$. Actually, $d = \min X_d$. – lhf May 16 '21 at 14:13
So you're saying there's a direct function from the additive order to the size of the partition? – nomad May 16 '21 at 14:46
Yes. See https://en.wikipedia.org/wiki/Euler%27s_totient_function – lhf May 16 '21 at 16:04
See also https://math.stackexchange.com/a/2782281/589 – lhf May 16 '21 at 20:48

Partition integer range (1 to n) as multiples of divisors of n

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