Algorithm to uniquely determine a number using two adjacent digits 
(Russia)
Arutyun and Amayak perform a magic trick as follows. A spectator writes down on a board a sequence of $N$ (decimal) digits. Amayak covers two adjacent digits by a black disc. Then Arutyun comes and says both closed digits (and their order). For which minimal N can this trick always work? NOTE:Arutyun and Amayak have a strategy determined beforehand.

I thought the question was somehow similar to this one, but the same idea doesn't seem to work.
I understand that it might be useful to look for an easy way to do the above magic trick(i.e. it does not require the magician to memorize every possible configuration rather the magician can assign some value to each N which will allow him to know which two numbers his friend covers).
The questions that bug me are:
1)Why must the covered digits be adjacent? Why is this necessary?
2)How can the magician know the order of the digits?
To answer these questions, I first defined $\overline{x_1\cdots x_N}$ to be the number written on the board. Then I considered the quantity $a=x_1x_2+x_2x_3+\cdots +x_{N-1}x_N$. This did not work out well for me.
Any help regarding this will be appreciated.
 A: 101 digits will be enough -- that allows 100 different positions of two adjacent digits, which allows Amayak to communicate the sums of the odd-position and even-position digits modulo 10. It is now easy for Arutyun to reconstruct the missing odd and the missing even digit.
On the other hand, 100 digits (or less) is not enough. There are only 99 choices for which two digits to cover, so only $99\cdot 10^{98}$ different states of the blackboard for Arutyun to observe when he returns, which is fewer than the $10^{100}$ sequences the spectator has to choose between.
A: 1)Why must the covered digits be adjacent? Why is this necessary?
Because, if you do not ask the covered digits to be adjacent, finding matching upper and lower bounds is a challenge (I do not know what is the minimal suitable value of N).
For instance, for $N = 20$, you can split the array of digits in two parts and, for each part, cover the digit #$k$ if the sum of the digits in the part is equal to $k$ mod 10. This is essentially the same method as above.
However, simple lower bounds only require that $100 \leq N(N-1)/2$, because Arutyun can see $N(N-1)/2 \cdot 10^{N-2}$ arrays and should be able to reconstruct $100^N$ arrays. This ensures that $15 \leq N$, but not that $20 \leq N$, so there is still an open gap about which is the smallest suitable value of $N$.
