I think it is hopeless to decide whether the number of digits of Graham's number is even or odd because the only way that I can think of is determining the logarithm with accuracy $0.1$ or even better, if the logaithm happens to be near an integer.
Is that right, or is there any trick to decide the question ?
The definition of Graham's number :
If we have the recursion $G_0=4$ , $G_{n+1}=3\uparrow^{G_n} 3$ for all $n\ge 0 $, where $a\uparrow^b c$ denotes Knut's up-arrow-notation, then Graham's number is $G_{64}$.
Knut's uparrow notation works as follows :
$a\uparrow b=a^b$
$a\uparrow \uparrow b=a\uparrow a\uparrow a\uparrow ...\uparrow a\uparrow a\uparrow a\ $ with $b$ $a's$
$a\uparrow \uparrow \uparrow b=a\uparrow \uparrow a \uparrow \uparrow ... \uparrow \uparrow a\uparrow \uparrow a\ $ with $b$ $a's$
and so on.
$a\uparrow^b c$ means $b$ up-arrows between $a$ and $c$
Note, that the calculation is done from right to left, so for example $3\uparrow 3\uparrow 3=3\uparrow 27=3^{27}$ and not $3\uparrow 3\uparrow 3=27\uparrow 3=27^3=3^9$