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

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Puzzle: One Four: $64$

Using exactly $1$ instance of the number $4$, the task is to write an expression for $64$, using whatever arithmetical operations you consider necessary.


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


We have that:

\(\ds 5^{32}\) \(=\) \(\ds 232 \, 83064 \, 36538 \, 69628 \, 90625\)
\(\ds 24!\) \(=\) \(\ds 6204 \, 48401 \, 73323 \, 94393 \, 60000\) Factorial of $24$
\(\ds 6^{32}\) \(=\) \(\ds 79586 \, 61109 \, 94640 \, 08843 \, 91936\)
\(\ds \leadsto \ \ \) \(\ds 5\) \(<\) \(\ds \sqrt [32] {24!}\)
\(\ds \) \(<\) \(\ds 6\)
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \floor {\surd \surd \surd \surd \surd \paren {4!} !}\) \(=\) \(\ds 5\)


Then:

\(\ds \floor {\surd \floor {\surd \surd \surd \surd \surd \paren {4!} !} !}\) \(=\) \(\ds \floor {\surd 5 !}\) from $(1)$
\(\ds \) \(=\) \(\ds \floor {\surd 120}\)
\(\ds \) \(=\) \(\ds \floor {10.95 ...}\)
\(\text {(2)}: \quad\) \(\ds \) \(=\) \(\ds 10\)


Then:

\(\ds \floor {\surd \floor {\surd \floor {\surd \surd \surd \surd \surd \paren {4!} !} !} !}\) \(=\) \(\ds \floor {\surd 10!}\) from $(2)$
\(\ds \) \(=\) \(\ds \floor {\surd 3628800}\)
\(\ds \) \(=\) \(\ds \floor {1904.94 ...}\)
\(\text {(3)}: \quad\) \(\ds \) \(=\) \(\ds 1904\)

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.

$\blacksquare$


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 $6$:

$6 = \floor {\surd \floor {\surd \floor {\surd \surd \surd \surd \surd \paren {4!} !} !} }!$


Sources

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