Four Fours/Lemmata/One Four/64/Solutions/1

Solution

 * $64 = \floor {\surd \surd \surd \surd \surd \surd \surd \surd \surd \floor {\surd \surd \surd \surd \surd \surd \surd \surd \surd \floor {\surd \surd \surd \surd \surd \surd \surd \surd \surd \surd \surd \surd \surd \floor {\surd \surd \surd \surd \surd \surd \surd \surd \floor { \surd \surd \surd \surd \surd \surd \surd \surd \surd \surd \surd \floor {\surd \floor {\surd \floor {\surd \surd \surd \surd \surd \paren {4!} !} !} !} !} !} !} !} !}$

where:
 * $x!$ denotes the factorial of $x$
 * $\floor x$ denotes the floor function of $x$.

Proof
Note that in the below, factorial has a higher binding priority than the square root function.

That is, $\sqrt x!$ means $\sqrt {\paren {x!} }$ and not $\paren {\sqrt x!}$.

We have that:

Then:

Then:

The pattern continues.

We see that the $k$th floor function evaluates to the integer $a_k$ such that:


 * ${a_k}^{2^t} < a_{k - 1} ! < \paren {a_k + 1}^{2^t}$

where $t$ is the number of square root signs between the floor function delimiters.

Laborious evaluation then allows us to construct the following table:


 * $\begin {array} {crrlll}

k & a_k & t & {a_k}^{2^t} & a_{k - 1} ! & \paren {a_k + 1}^{2^t} \\ 0 & 24  &    &   &   &   \\ 1 & 5    & 5  & 0.23 \times 10^{23} & 0.62 \times 10^{24} & 0.79 \times 10^{25} \\ 2 & 10  & 1  & 100 & 120 & 121 \\ 3 & 1904 & 1  & 36 \, 25216 & 36 \, 28800 & 39 \, 29025 \\ 4 & 442  & 11 & 0.67 \times 10^{5417} & 0.42 \times 10^{5419} & 0.68 \times 10^{5419} \\ 5 & 6673 & 8 & 0.1062 \times 10^{979} & 0.1097 \times 10^{979} & 0.1104 \times 10^{979} \\ 6 & 577 & 13 & 0.4 \times 10^{22619} & 0.9 \times 10^{22623} & 0.5 \times 10^{22625} \\ 7 & 422 & 9  & 0.14 \times 10^{1344} & 0.25 \times 10^{1344} & 0.49 \times 10^{1344} \\ 8 & 64  & 9  & 0.58 \times 10^{924} & 0.21 \times 10^{926} & 0.16 \times 10^{928} \\ \end {array}$

From Floor of Root of Floor equals Floor of Root, we have:
 * $\ds \floor {\sqrt {\floor x} } = \floor {\sqrt x}$

so we need only to take the floor function of the entire result, and before the factorial is evaluated.

The result follows.

Note in passing that $\floor {\surd 10} = 3$, allowing us an expression for $3$ using one $4$:


 * $3 = \floor {\surd \floor {\surd \floor {\surd \surd \surd \surd \surd \paren {4!} !} !} }$

from $(2)$ above.

Similarly, as $3! = 6$, we have a similar expression for $6$:
 * $6 = \floor {\surd \floor {\surd \floor {\surd \surd \surd \surd \surd \paren {4!} !} !} }!$