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, Exq({1,2,4}) = 10, and Exq({3,4,5,6}) = 42, 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 Exq(A) = 1, otherwise Exq(A) = 0.

If  A = {1,2}, then 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, Exq(A) ≤ 4.

Let us assume that 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 values were negative, then addition would not produce a positive integer.  If exactly one value were negative, then multiplication would not produce a positive integer.  So both values must be positive.  And if the two values were equal, then subtraction would not produce a positive integer.  

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 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 ≥ 2 and b ≥ 3.  But this implies a sum of at least 5, too large to obtain Exq(A) = 4 with just four operations.  Therefore, Exq(A) ≤ 3.

Back to Theorems and Conjectures.