What is the *middle* digit of $3^{100000}$?

Question

The decimal representation of $3^{100000}$ has $47713$ digits. What is the $23857^{th}$ digit - i.e. the one in the $10^{23856}$'s place?

There are lots of questions on this site asking for the last digit of various large exponents, which are well handled by modular arithmetic. The last digit of $3^{100000}$ is $1$ by such methods. There are some fewer number of questions asking for the first digit of a number, which can be handled by considering the fractional part of the base ten logarithm of the number. The first digit of the number is $1$.

Either method can be modified to give the first few or last few digits, but they don't generalize well beyond that, and there's no obvious way to combine the methods.

Is there a method to find arbitrary digits of small integers raised to large ones? Is there any reason to suspect that computing a digit in the middle is, in terms of computational complexity, a more difficult problem than computing digits on the edge?

Answer 1

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:

• ON THE CONSISTENCY AND INDEPENDENCE OF SOME SET-THEORETICAL AXIOMS

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:

• Meelo.dpr

And here is a link to the number $3^{100000}$ itself, which is too large to fit into MSE's margins:

• drie5.txt

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:

The purpose of the $\sf ZFC$ Axiom of Infinity

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.

Answer 2

I wondered how Lucian could come to the correct result within one hour after the question appeared. So I gave Mathematica a chance. Here is the output:

Answer 3

Aside from special cases where 'tricks' might aid in solving a specific answer, I suspect finding an arbitrary digit or slice of digits roughly reduces to the time complexity of base conversion. For example, if you want to know the middle (and only middle!) digit of a number with an odd number of digits, there may in fact be a special trick through modular arithmetic to get the answer quickly. I don't know of one, maybe Lucian does.

However, if you want to know an arbitrary digit of $N=b^k$, I believe the answer to your question is similar to knowing the complexity of base conversion from base b to base 10.

Rough argument

Storage of $N=b^k$ in base 10 will require $ \lceil \log_{10}(N)\rceil=\lceil k*\log_{10}(b)\rceil$ digits.

Computing N by repeated squaring will require $\lceil \log_2(k) \rceil$ multiplication operations each of which involves updating the $\lceil k*\log_{10}(b) \rceil$ elements of the representation.

Therefore, it appears you're bounded by $ O(k*log_2(k))$ operations as $k$ increases.

This differs by a factor of $k$ from modular arithmetic tricks which will generally get you the last digits in $O(log_2(k))$ time (i.e., if you use the same repeated squaring approach).

So where is the difference?
The 'trick' with first/last digits is to restrict the representation to a fixed constant. Although $k$ may become larger in $b^k$, the representation of your answer does not have to grow, so first/last digits can be computed efficiently.

So I think the real underlying question is:

Can arbitrary digits of $N=b^k$ be determined using a representation of fixed width (fixed number of bits)?

Answer 4

Using a computer program (that just calculates the whole number and extract the required digit), I found that the answer is 7.

This is the program (which is in Haskell):

digits :: Integer -> [Int]
digits = map (read . (:[])) . show

main = print ((digits (3^100000)) !! 23857)