Does every Integermania problem have a solution?

Theorem 3:  There exists an Integermania solution, without rounding, for every integer n and every non-empty set of integers A.

Proof:  Suppose A = {1}.  If n percent signs are used, we have .  If a negative integer is desired, take opposites.  Therefore every Integermania problem can be solved with the set A = {1}.

Since 0! = 1, and Γ(2) = 1, there exists a function such that when a is 0 or 2.  Then , so every Integermania problem can be solved with the set A = {0} or the set A = {2}.

Now suppose A = {a}.  The function d(a) is the number of divisors of the positive integer a.  For .  This implies that there exists some integer number of compositions of the functions d(a) for which .  Therefore, for any non-negative integer a, there exists a function such that .  And since a can be turned into its opposite by a negative sign, exists with for every integer.  Thus every Integermania problem can be solved for any integer a, positive, negative, or zero, in the singleton set A = {a}.

Now suppose .  If percent signs are used, we have .  Therefore every Integermania problem can be solved with any set A of integers.

Comments:  This existence proof used level 6 operations, and by no means gives the most exquisite solutions.  All of the sets used on this site can be solved with level 4 operations or below, although the unary surcharges will cause solutions to have higher exquisiteness levels.  All of the examples below have n percent signs.

First Four Composites:
First Four Naturals:
First Four Primes:
Four Fours:
JCCC Letters:
JCCC Zips:

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