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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
\begin{equation} -\log\sqrt{1\% \cdots \%} = -\log(10^{-2n})^{1/2} = n \end{equation}If a negative integer is desired, take opposites. Therefore every Integermania problem can be solved with the set $A = \{1\}$.
Since $0! = 1$, and $\Gamma(2) = 1$, there exists a function $g_a(x)$ such that $g_a(a) = 1$ when $a$ is 0 or 2. Then
\begin{equation} -\log\sqrt{g_a(a)\% \cdots \%} = -\log(10^{-2n})^{1/2} = n \end{equation}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 $a \ge 3$, then $2 \le d(a) < a$. This implies that there exists some integer number of compositions of the functions $d(a)$ for which $d(d( \cdots d(a) \cdots )) = 2$. Therefore, for any non-negative integer $a$, there exists a function $g_a(x)$ such that $g_a(a) = 1$. And since $a$ can be turned into its opposite by a negative sign, $g_a(x)$ exists with $g_a(a) = 1$ 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 $A = \{a_1, a_2, ..., a_k \}$. If $n + \sum\limits_{i=2}^k |a_i|$ percent signs are used, we have
\begin{equation} -\log\sqrt{g_{a_1}(a_1)\% \cdots \%} - \sum\limits_{i=2}^k |a_i| = -\log\left( 10^{-2\left( n + \sum\limits_{i=2}^k |a_i| \right)} \right)^{1/2} - \sum\limits_{i=2}^k |a_i| = n \end{equation}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.
| Dave's Birthday | $8 \times (7-7) - \log\sqrt{1 \% \cdots \%} = -\log(10^{-2n})^{1/2} = n$ |
| First Four Composites: | $-\log\sqrt{(8+6-9-4) \% \cdots \%} = -\log(10^{-2n})^{1/2} = n$ |
| First Four Naturals: | $(4-3-2) \times \log\sqrt{1 \% \cdots \%} = -\log(10^{-2n})^{1/2} = n$ |
| First Four Nonsquares: | $(5 - 6) \times \log\sqrt{(3 - 2) \% \cdots \%} = -\log(10^{-2n})^{1/2} = n$ |
| First Four Odds: | $(7-3-5) \times \log\sqrt{1 \% \cdots \%} = -\log(10^{-2n})^{1/2} = n$ |
| First Four Primes: | $-\log\sqrt{(3 \times 5 - 2 \times 7) \% \cdots \%} = -\log(10^{-2n})^{1/2} = n$ |
| Four Fours: | $4 - 4 - \log\sqrt{\dfrac44 \% \cdots \%} = -\log(10^{-2n})^{1/2} = n$ |
| Four Nines: | $9 - 9 - \log\sqrt{\dfrac99 \% \cdots \%} = -\log(10^{-2n})^{1/2} = n$ |
| JCCC Letters: | $-\log\sqrt{(10 - 3 - 3 - 3)\% \cdots \%} = -\log(10^{-2n})^{1/2} = n$ |
| Largest Four Digits: | $(8 - 9) \times \log\sqrt{(7 - 6) \% \cdots \%} = -\log(10^{-2n})^{1/2} = n$ |
| Mile and a Foot: | $(5 + 2 - 8) \times \log\sqrt{1 \% \cdots \%} = -\log(10^{-2n})^{1/2} = n$ |
| Quattro Ones: | $1 \times 1 - 1 - \log\sqrt{1 \% \cdots \%} = -\log(10^{-2n})^{1/2} = n$ |
| Ralph's Birthyear: | $(8 - 9) \times \log\sqrt{1^4 \% \cdots \%} = -\log(10^{-2n})^{1/2} = n$ |
| Ramanujan: | $\dfrac{7 - 9}{2} \times \log\sqrt{1 \% \cdots \%} = -\log(10^{-2n})^{1/2} = n$ |
| Zip 66210: | $0 \times (6 + 6 + 2) -\log\sqrt{1 \% \cdots \%} = -\log(10^{-2n})^{1/2} = n$ |
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