8.82 Suppose there are 128 ones written on the blackboard. In each step, you can erase any two numbers $a$ and $b$ on the blackboard, and write $ab+1$. After 127 such steps, only one number remains. Let the maximum possible value of this remaining number be $A$. Find the last digit of $A$.


[Solution] First, we prove that as long as each step involves the smallest two numbers on the blackboard, the final remaining number can reach its maximum value. We denote the operation on numbers $a$ and $b$ on the blackboard as $a * b = ab + 1$. Suppose at some step, the operation is not performed on the smallest two numbers $x$ and $y$, then before $x$ and $y$ "meet," they have already "met" with some $a_{i}, b_{j}$, and these $a_{i}, b_{j}$ are all greater than $x$ and $y$, i.e.,
\[
\begin{aligned}
a & =\left(\left(\cdots\left(\left(x * a_{1}\right) * a_{2}\right) \cdots\right) * a_{k}\right) \\
& =a_{1} a_{2} \cdots a_{k}\left(x+\frac{1}{a_{1}}+\frac{1}{a_{1} a_{2}}+\cdots+\frac{1}{a_{1} a_{2} \cdots a_{k}}\right) \\
& =a_{2} a_{3} \cdots a_{k}\left(a_{1} x+1+u\right), \\
b & =\left(\left(\cdots\left(\left(y * b_{1}\right) * b_{2}\right) \cdots\right) * b_{n}\right) \\
& =b_{1} b_{2} \cdots b_{n}\left(y+\frac{1}{b_{1}}+\frac{1}{b_{1} b_{2}}+\cdots+\frac{1}{b_{1} b_{2} \cdots b_{n}}\right) \\
& =b_{1} b_{2} \cdots b_{n}(y+v),
\end{aligned}
\]

where $u > 0, v > 0$ represent the sum of the remaining numbers in the respective parentheses. We can assume that in (2) there is $y < x$, and thus
\[
\begin{aligned}
a * b & = a_{2} a_{3} \cdots a_{k} b_{1} b_{2} \cdots b_{n} \left( (a_{1} x + 1 + u)(y + v) + 1 \right) \\
& = a_{2} a_{3} \cdots a_{k} b_{1} b_{2} \cdots b_{n} \left( a_{1} x y + a_{1} x v + y + v + u y + u v + 1 \right) \\
& = a_{2} a_{3} \cdots a_{k} b_{1} b_{2} \cdots b_{n} \left( a_{1} x y + a_{1} x v + y + v + u y + u v + 1 \right) \\
& > a_{2} a_{3} \cdots a_{k} b_{1} b_{2} \cdots b_{n} \left( a_{1} x y + y + 1 \right) \\
& = (a_{1} x + 1) * (y + 1).
\end{aligned}
\]

This shows that the exchange results in an increase in the final remaining number. This proves the initial claim.

Thus, as long as each operation is performed on the smallest two numbers on the blackboard, the maximum possible value can be obtained. Therefore, after the first 64 steps, there are 64 threes on the blackboard; after another 32 steps, there are 32 tens on the blackboard; continuing this process, the last remaining number will have a units digit of 2.