Raising Squares Part 3

I used Python code for a solution, because I'm the laziest person in the world. But this can be approached logically in a simple way. Intuition says we probably want larger numbers closer "to the ground," as it were. We want to keep them out of the higher boxes. Let's not use this rule rigorously, but we can keep it in the back of our minds. Then, we remember that $1^x=1 \forall x \in \mathbb{R}$ ! So, we can actually put $1$ in one of the bottom boxes, and rid our set of those pesky fellows $5$ and $6$ , right in the boxes above $1$ . So, now the answer is of the form $1+a^{b^c}$ , where we have $a,b,c \in [2,3,4]$ . We only have $3!=6$ cases to check, and they're rather easy to sort manually. Remember, we want higher numbers closer to the "ground." But clearly, $4^{3^2}>4^{2^3}$ , and $4^{2^3}=2^{16}$ . Moving onto the next cases, we have $2^{3^4} = 2^{81} > 2^{16}$ and $2^{4^3} = 2^{64} > 2^{16}$ . The last cases, conveniently wrapped into one: $3^{2^4}=3^{4^2}=3^{16}>2^{16}$ . Therefore $2^{16}=65536$ is clearly the minimum for what we had left, and so the answer is $\boxed{65537}$ .

This is awfully informal, but eh.

Consider a single tower $\square^{\square^\square}$ . If one of the numbers is $1$ , then to achieve the minimum, $1$ must be at the bottom: $1^{\square^\square} = 1 < 2 \le \square_1 \le (\square_1)^{\square^\square}$ for any $\square_1 \neq 1$ . Thus we may assume that $1$ occupies the leftmost square: we know $1$ is at the base, and addition is commutative. Thus now we have $1^{\square^\square} + \square^{\square^\square}$ .

Next, consider a tower $\square^{\square^\square}$ . Take any of the squares as the variable, and fix the other two. Observe that the resulting function in one variable is strictly increasing on the positive integers: for example, $x \mapsto x^{\square^\square}$ is strictly increasing when $x$ is a positive integer, for any positive integers filling the squares.

This essentially means that if we can replace a square with a smaller number (and retaining that there is no repetition, perhaps by changing the numbers present in the tower with the $1$ ), the current arrangement is not the smallest. In other words, if there exists a $5$ or a $6$ in the second tower, we know that we can replace it by some number in $2,3,4$ that is not present in the tower (and replace the corresponding number in the first tower with the replaced number). For example, $1^{3^6} + 2^{4^5}$ can be replaced to $1^{5^6} + 2^{4^3}$ . Thus we can assume that the second tower only has $2,3,4$ . In addition, this means the $5,6$ must go to the first tower, and their locations are actually not important since either way the left tower is $1$ raised to something, giving $1$ .

Finally, the $6$ possibilities are actually pretty easy to compare:

• Since $3^4 > 4^3$ , $2^{3^4} > 2^{4^3}$ . This rules out $2^{3^4}$ . The one remaining is $2^{4^3} = 2^{64}$ .
• Since $2^4 = 4^2$ , $3^{2^4} = 3^{4^2}$ . We can just take one of them, $3^{2^4} = 3^{16}$ .
• Since $2^3 < 3^2$ , $4^{2^3} < 4^{3^2}$ . This rules out $4^{3^2}$ . The one remaining is $4^{2^3} = 4^8 = 2^{16}$ .
• Since $64 > 16$ , $2^{64} > 2^{16}$ . This rules out $2^{4^3}$ .
• Since $3 > 2$ , $3^{16} > 2^{16}$ . This rules out both $3^{2^4}$ and $3^{4^2}$ .

Thus the second tower should be $4^{2^3}$ to minimize the result. We thus have $1^{\square^\square} + 4^{2^3} = 1 + 4^8 = \boxed{65537}$ . We also know that there are exactly $4$ permutations giving this result (the $1$ can go in two places; the $5$ can go to either place in the same tower as the $1$ ; the rest is fixed).

1^(5^6) + 4^(2^3) OR 1^(6^5) + 4^(2^3) = 65537

1^(5^6) + 4^(2^3) = 65537

1^(5^6) + 4^(2^3) OR 1^(6^5) + 4^(2^3) = 65537