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What is the maximum Integermania exquisiteness of a set of $n$ values?

Theorem 2: In an Integermania problem where set $A$ has $n$ values, the level 1 exquisiteness of $A$ will be less than or equal to 1 when   $n=1$,   and less than or equal to 3 when   $n=2$.   See proof below.

Comments: The maximum exquisiteness of larger sets is unknown. However,   $\operatorname{Exq}(\{1,2,4\}) = 10$,   and   $\operatorname{Exq}(\{2,3,4,22\}) = 52$,   so these are lower bounds to the maximums.

Proof: A singleton set does not use binary operations, and hence can only produce itself. Therefore, if   $a=1$,   then $\operatorname{Exq}(A) = 1$,   otherwise   $\operatorname{Exq}(A) = 0$.

If   $A = \{1,2\}$,   then   $\operatorname{Exq}(A) = 3$.   The theorem claims that this set reaches maximum exquisiteness, although this is not the only doubleton set with exquisiteness 3.

Doubleton sets   $A = \{a,b\}$   will use exactly one of the four available binary operations. Addition and multiplication are commutative. Subtraction and division will each produce at most one positive integer. Therefore,   $\operatorname{Exq}(A) \le 4$.

Let us assume that   $\operatorname{Exq}(A) = 4$,   and find a contradiction. We expect to find values $a$ and $b$ for which the sum, difference, product, and quotient give the numbers 1, 2, 3, and 4 (but not necessarily in that order).

• If either $a$ or $b$ were zero, then multiplication would not produce a positive integer.
• If both $a$ and $b$ were negative, then addition would not produce a positive integer.
• If exactly one value of $a$ and $b$ was negative, then multiplication would not produce a positive integer.
• If the two values $a$ and $b$ were equal, then subtraction would not produce a positive integer.

So both values $a$ and $b$ must be positive, and not equal to one another.

Since the values must be different, let us assume that   $a < b$. Since both   $b+a$   and   $b-a$   must be positive integers, we can add and subtract these to find that $2b$ and $2a$ are both positive integers, therefore $a$ and $b$ are both multiples of one-half. If both were odd multiples, then multiplication would not produce a positive integer. If exactly one was an odd multiple, then addition would not produce a positive integer. Therefore, both must themselves be positive integers (even multiples of one-half).

If   $a=1$   (the smallest positive integer), then multiplication and division would produce the same positive integer. Therefore,   $a \ge 2$   and   $b \ge 3$.   But this implies a sum of at least 5, too large to obtain   $\operatorname{Exq}(A) = 4$   with just four operations. Therefore,   $\operatorname{Exq}(A) \le 3$.

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