Remarkable subsets

Let $b$ be the smallest positive element in $A$ . Then $-b \in A$ by rule 1, and by rule 3, all $kb \in A$ for all odd $k$ .

There are now two possibilities:

• $0 \in A$ , in which case (by rule 2 again) $kb \in A$ for all $k$ : $A \supseteq \{\dots, -4b, -3b, -2b, -b, 0, b, 2b, 3b, 4b, \dots \}$

• $0 \not\in A$ , in which case $kb \in A$ for all odd $k$ . $A \supseteq \{\dots, -5b, -3b, -b, b, 3b, 5b, \dots \}$

In fact, equality holds in either case: set $A$ contains no other elements. Proof: if $a \in A$ is not an odd multiple of $kg$ , use rule 3 repeatedly to add or subtract $2b$ until you find a number $a'$ strictly between $-b$ and $+b$ . If $a' < 0$ , swap it out for $-a' > 0$ , which must also be an element by rule 1. Then either $a' = 0$ , showing that $a = kb$ for even $k$ ; or $0 < a < b$ , which contradicts the assumption that $b$ is minimal.

Now use rule 2 to conclude that $N$ should be a multiple of the "step size" in $A$ : $b | N$ (first case) or $2b | N$ (i.e. $b | \tfrac12N$ , provided $N$ is even) (second case). In the first case we must rule out $b = 1$ because then $A = \mathbb Z$ trivially. For $N = 18$ we find

• five sets of the first kind: $b = 2, 3, 6, 9, 18$ $A_1 = \{\dots, -8, -6, -4, -2, 0, 2, 4, 6, 8, \dots \} \\ A_2 = \{\dots, -12, -9, -6, -3, 0, 3, 6, 9, 12, \dots \} \\ A_3 = \{\dots, -24, -18, -12, -6, 0, 6, 12, 18, 24, \dots \} \\ A_4 = \{\dots, -36, -27, -18, -9, 0, 9, 18, 27, 36, \dots \} \\ A_5 = \{\dots, -72, -54, -36, -18, 0, 18, 36, 54, 72, \dots \}$

• three sets of the second kind: $b = 1, 3, 9$ $A_6 = \{\dots, -7, -5, -3, -1, 1, 3, 5, 7, \dots \} \\ A_7 = \{\dots, -21, -15, -9, -3, 3, 9, 15, 21, \dots \} \\ A_8 = \{\dots, -63, -45, -27, -9, 9, 27, 45, 63, \dots \}$

The answer to the problem is: there exist $\boxed{\text{eight}}$ remarkable sets of type $18$ .

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In general, the number of sets $A$ of the given description is $n(N) = (\tau(N) - 1) + \tau(\tfrac12 N),$ where $\tau(x)$ is the number of positive divisors of $x$ . Given the prime factor decomposition $N = 2^{e_2}\cdot 3^{e_3}\cdot 5^{e_5} \cdots p^{e_p}\cdots,$ this becomes $n(N) = (2e_2+1)\cdot (e_3 + 1)\cdot (e_5 + 1)\cdots(e_p+1)\cdots - 1.$ (Check: if $N = 18$ then $e_2 = 1, e_3 = 2$ , and $n(N) = (2+1)\cdot (2+1) - 1 = 3\cdot 3 - 1 = 8$ .)

Arjen has the brilliant solution, but I thought I would post the result of my little script

first column is the generator, following values are elements of the set between 0-18. I added 0 as optional value for odd generators, even generators already produce 0

0 0 18

1 1 3 5 7 9 11 13 15 17

1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

2 0 2 4 6 8 10 12 14 16 18

3 3 9 15

3 0 3 6 9 12 15 18

6 0 6 12 18

9 9

9 0 9 18

options: 9

This includes the trivial set Z, so the solution is 8