The Impossibly Long... Blackboard

Step 1:

The number of decimal digits of the integers from 1 to $10^{100}-1$ is

$S = \sum_{k=1}^{100} (9k \times 10^{k})$

This is due to there being 9 integers from 1 to 9 (1 digit), 90 integers from 10 to 99 (2 digits), 900 integers from 100-999 (3 digits), etc.

Because $10^{100}$ has 101 digits, $100T = 101+S$ .

Step 2:

Now, consider the product

$\prod_{k=1}^{10^{100}} (a_{k}+1)$

where $\lbrace a_{1}, a_{2}, \ldots \rbrace$ is a permutation of the integers from 1 to $10^{100}$ . Note that this is an invariant under iteration of $a+ab+b = (a+1)(b+1)-1$ .

Since this product is $(10^{100}+1)!$ at the beginning, the number at the end will be $(10^{100}+1)!-1$ , our $n = 10^{100}$ . Thus, $n = 10^{100}$ and $\log_{10}(n) = 100$ .

Step 3:

It is then obvious that $\lfloor T \rfloor \pmod{100}$ is asking for the digits in the hundreds and thousands place of $100T = 101+S$ . To find this, we must evaluate $T \pmod{10000}$ . Since

$S = \sum_{k=1}^{100} (9k \times 10^{k}) \equiv \sum_{k=1}^{3} (9k \times 10^{k}) \equiv 8890 \pmod{10000}$

we can therefore conclude

$\lfloor T \rfloor = \lfloor \frac{101+S}{8890} \rfloor \equiv \lfloor \frac{101+8890}{100} \rfloor \equiv \boxed{89} \pmod{100}$