Subtiles 2

This month's puzzle is straightforward in comparison to January's puzzle. The first thing to find was the upper bound of the $K$-ominos. Given our constraints and a $13\times 13$ board, the max $K$-omino bound was $17$.

• max $K=17$: $\frac{17\cdot (17+1)}{2} = 153\leq 169$

This is only an upper bound value. Therefore, each equation must yield an answer between $1$ to $17$. When looking through the equations, there were obvious constraints for each constant.

$b^2$ and $(b-1)^2$

$b^2\in \{1,...,17\}$ and $(b-1)^2\in \{1,...,17\}$. Given that $b^2=n$ and $(b-1)^2=m$, then

• $b^2-(b-1)^2=2b-1=m-n$
• $b=\frac{m-n+1}{2}=\pm m$
• This forces $m$ to be a perfect square, then $b$ is an integer
• $b^2\leq 16$ then $b \in \{-4,-3,-2,-1,1,2,3,4\}$
• $(b-1)^2 \geq 1$ so $b\neq 1$
• $(b-1)^2 \leq 16 \rightarrow |b-1|\leq 4$ so $b\in \{-3,...5\}$
• Thus $b\in \{-3,-2,-1,2,3,4\}$

$\frac{\sqrt{a+2}}{a}$

Given that $\frac{\sqrt{a+2}}{a}\in \mathbb{Z}$ and $\frac{\sqrt{a+2}}{a} \in \{1,2,...,17\}$.

• $\sqrt{a+2} = ak \rightarrow a+2 =a^2k^2$
• $a^2k^2-a-2=0$
• $a = \frac{1\pm \sqrt{1+8k^2}}{2k^2}$
• $1+8k^2$ must be a perfect square for $a$ to be rational
• Checking values $1$ to $17$, the only $k$ that work are $k=1,6$ where $a=2$ and $\frac{1}{4}$, respectively
• This narrows the search space a lot

Eliminating $a = 2$

• We need $c > a$ for $\sqrt{c-a}$ to be real and nonzero (expression 11)
• We also need $\log_c(a)$ to be a positive integer $n$, which means $a = c^n$
• If $a = 2$, then $c^n = 2$ with $c > 2$
• But $c > 2$ and $n \geq 1$ gives $c^n > 2$, so $c^n = 2$ is impossible
• Therefore $a = 2$ is ruled out, leaving:

$a = \frac{1}{4}$

Narrowing $b$ with $a = \frac{1}{4}$

• The expression $4a - 5b$ must be a positive integer
• Substituting $a = \frac{1}{4}$: $4\cdot\frac{1}{4} - 5b = 1 - 5b$
• For this to land in $\{1,...,17\}$ we need $b \leq 0$
• This eliminates $b \in \{2, 3, 4\}$, leaving:

$b \in \{-3, -2, -1\}$

Pinning $c$ from $c + 2a$

• The expression $c + 2a = c + \frac{1}{2}$ must be a positive integer $m$, so:
  • $c = m - \frac{1}{2}$ for $m \in \{1, 2, ..., 17\}$
• The constraint $c > a = \frac{1}{4}$ requires $m \geq 1$
• $c \neq 1$ requires $m \neq \frac{3}{2}$ (always satisfied since $m$ is an integer)
• So $c \in \{\frac{1}{2}, \frac{3}{2}, \frac{5}{2}, ...\}$

Eliminating $b = -1$

• $c^b = c^{-1} = \frac{1}{c}$ must be a positive integer, so $c = \frac{1}{j}$ for some positive integer $j$
• From above, $c = m - \frac{1}{2}$, meaning $\frac{1}{j} = m - \frac{1}{2}$
• The only solution with $m$ a positive integer and $j$ a positive integer is $m = 1, j = 2$, giving $c = \frac{1}{2}$
• $(b+c)/(c-1) = (-1 + \frac{1}{2})/(\frac{1}{2} - 1) = (-\frac{1}{2})/(-\frac{1}{2}) = 1$
• $b^2 - b/c = 1 - (-1)/(\frac{1}{2}) = 1 + 2 = 3$
• $(b+9)/\sqrt{c-a} = 8/\sqrt{\frac{1}{4}} = 8/\frac{1}{2} = 16$
• $(a^b - 4)/(6c+1) = (4 - 4)/4 = 0$
• The expression $(a^b-4)/(6c+1) = 0$ is not a positive integer, so $b = -1$ fails

Eliminating $b = -2$

• $c^b = c^{-2} = \frac{1}{c^2}$ must be a positive integer, so $c^2 = \frac{1}{j}$ giving $c = \frac{1}{\sqrt{j}}$
• Combined with $c = m - \frac{1}{2}$: $m - \frac{1}{2} = \frac{1}{\sqrt{j}}$
• For $m = 1$: $\frac{1}{\sqrt{j}} = \frac{1}{2}$, so $j = 4$, giving $c = \frac{1}{2}$
• $(a^b - 4)/(6c+1) = (({\frac{1}{4}})^{-2} - 4)/(3+1) = (16-4)/4 = 3$
• $(b^3 + 2c)/(b+2c) = (-8 + 1)/(-2 + 1) = -7/-1 = 7$
• $6c - 4b = 3 + 8 = 11$
• $\log_c(a) = \log_{1/2}(1/4) = 2$
• $b/(a-1) = -2/(1/4 - 1) = -2/(-3/4) = 8/3$
• The expression $b/(a-1) = \frac{8}{3} \notin \mathbb{Z}$, so $b = -2$ fails

$b = -3$ and $c = \frac{1}{2}$

• With $b = -3$ and $a = \frac{1}{4}$, $c^b = c^{-3}$ must be a positive integer, so $c = j^{-1/3}$ for positive integer $j$
• Combined with $c = m - \frac{1}{2}$, trying $m = 1$ gives $c = \frac{1}{2}$ and $c^{-3} = 8$
• Checking $b/(a-1) = -3/(1/4 - 1) = -3/(-3/4) = 4$
• All 37 expressions now evaluate to positive integers:

$\boxed{a = \tfrac{1}{4}, \quad b = -3, \quad c = \tfrac{1}{2}}$

K-omino Distribution#

Given that $12$ was the most constrained $k$-omino. There were only around 5 configurations that made sense. Building from the $12$ $k$-omino, I programmatically found the grid. I was able to solve this months puzzle to get the answer 9072.