In contrast with other programming languages from its time, Pascal (the programming language)
also supports a set type, implemented as a bit pattern. The bit patterns associated with
the Pascal set type are $256$ bits wide, but this limitation is not essential and can been replaced
with other (larger) values eventually.
See Wikipedia for a reference.
A rather detailed description of the set type implementation can be found
as well .
So we have the following practice:
- a bit pattern in a computer is a set type
We also know that
- a bit pattern in a computer is a natural number type
Indeed, everybody knows that a natural number can be represented as a binary i.e. a bit pattern.
The word "type" has been employed here in order to avoid confusion with other
(i.e the standard mathematical) "set" and "number" definitions.
More precisely: the
hereditarily finite sets
are in one-to-one correspondence with the natural numbers. And the latter fact is independent of
computers.
Examples.
$$
\begin{array}{l}
0 = 000 = \{\} \\
1 = 001 = \{0\} = \{\{\}\} \\
2 = 010 = \{1\} = \{\{\{\}\}\} \\
3 = 011 = \{0\; 1\} = \{\{\}\{\{\}\}\} \\
4 = 100 = \{2\} = \{\{\{\{\}\}\}\} \\
5 = 101 = \{0\; 2\} = \{\{\}\{\{\{\}\}\}\} \\
6 = 110 = \{1\; 2\} = \{\{\{\}\}\{\{\{\}\}\}\} \\
7 = 111 = \{0\; 1\; 2\} = \{\{\}\{\{\}\}\{\{\{\}\}\}\} \\
\cdots
\end{array}
$$
The above is related to the following reference, by Alexander Abian and Samuel LaMacchia:
If the curly brackets $\{\}$ are replaced by square brackets $\left[\,\right]$ then another
important fact is observed:
- a set type is is a natural number type is a sorted natural numbers array type
If now we devise
an equivalent of the elementary number operations with sorted arrays,
then we have virtual unlimited precision at our disposal. That this approach indeed works, shall be demonstrated at hand of the OP's question.
Here is a link to the complete (Delphi Pascal) program that does the job:
And here is a link to the number $3^{100000}$ itself, which is too large to fit into
MSE's margins:
The screen output of the program is the number of digits, the first, the middle and the last digit:
47713
1 2 1
So, indeed, as
Lucian says in a comment: it's "obvious" that the digit in the middle is a $\large\, 2$ .
Note. A power like $\,3^{100000}\,$ sounds quite impressive, but with a smart
algorithm, the number of operations is only $\,\ln_2(100000)\approx17$ . For
real numbers $\,x\,$ and a natural $\,n\,$ it goes as follows:
function power(x : double; n : integer) : double;
var
m : integer;
p, y : double;
begin
m := n; y := x; p := 1;
while m > 0 do begin
if (m and 1) > 0 then p := p * y;
m := m shr 1; { m := m / 2 }
y := y * y;
end;
power := p;
end;
Wikipedia reference:
Efficient computation with integer exponents .
BONUS. In view of the above, the following answer is interesting:
Reference is made to a
paper by Kaye and Wong,
where on page 499 we read:
It was observed in 1937 by Ackermann [1] that $\mathbb{N}$ with the membership relation
defined by
$n \in m$ iff the $n$th digit in the binary representation of $m$ is $1$
satisfies ZF$-$inf. This interpretation, formalized in ZF with $\omega$ in place of $\mathbb{N}$
yields a bijection between $\omega$ and the collection $V_\omega$ of hereditarily finite sets.
The last (decimal) digit of a number is the number's residue modulo $10$, so the general kind of problem you need to solve is
Given a power tower $a_0^{a_1^{\vdots^{a_n}}}$ and a small modulus $m$, compute $$a_0^{a_1^{\vdots^{a_n}}} \bmod m$$
The first step is of course to rewrite this to
$$(a_0\bmod m)^{a_1^{\vdots^{a_n}}} \bmod m$$
using general facts about modular arithmetic. Now in the particular case that $a_0$ is coprime to $m$, Euler's theorem allows us to rewrite this to
$$ (a_0\bmod m)^{\bigl(a_1^{\vdots^{a_n}} \bmod \varphi(m)\bigr)} \bmod m$$
in which the exponent is now a simpler instance of the original problem -- simpler because the tower is one level shorter and $\varphi(m)$ is less than $m$ (unless $m=1$ in which case anything mod $m$ is $0$ anyway and you can throw the entire tower away).
Once you have reduced the exponent you can raise $a_0\bmod m$ to it by standard techniques such as exponentiation by squaring -- or just by winging it if $m$ is as small as 10 and you have 32-bit arithmetic.
If $a_0$ and $m$ are not coprime, then the slick theoretical approach is to factor $m$ and use the Chinese remainder theorem, but a kind of poor man's shortcut is this bastard offshot of Euler's theorem:
If $a$ and $m$ are arbitrary positive integers and $b\ge\varphi(m)$, then
$$ a^b \equiv a^{\varphi(m)+(b\bmod\varphi(m))} \pmod m$$
So if you can just recognize whether or not the upper tower is large (which is easy), you can reduce the task to raising to a number that is not larger than $2\varphi(m)$.
Since your starting $m$ is $10$, you don't need any fancy machinery for computing $\varphi(m)$ -- just hardcode a table for $m$ up to $10$. (In fact the only $m$ you will need are $10,4,2,1$).
Best Answer
Hint:
Working in modulo $10$,
$$29583 \equiv 3$$
Thus,
$$(29583)^2 \equiv 9 \equiv -1$$