$(a)$ Let us construct $B$ from the right. Let $B_n$ and $A_n$ denote the $n^{th}$ digits of $B, A$ from the right.Choose $B_1=10-A_1$ and $B_n=9-A_n$ for all $n\ge 2$.

a) is obviosly b) if last digit is $x$, then exist super number B if and omly if $(x,10)=1$. c) It is not true Consider $A_n=2^{4*5^n}, B_n=5^{2^{n-1}}$. Obviosly $A_{n+m}=A_n\mod 10^n$ and $B_{n+m}=B_n \mod 10^n$ and $A_n*B_n=0\mod 10^n$.

Looks like $p$-adic numbers .

Amir Hossein wrote: Super number is a sequence of numbers $0,1,2,\ldots,9$ such that it has infinitely many digits at left. For example $\ldots 3030304$ is a super number. Note that all of positive integers are super numbers, which have zeros before they're original digits (for example we can represent the number $4$ as $\ldots, 00004$). Like positive integers, we can add up and multiply super numbers. For example: \[ \begin{array}{cc}& \ \ \ \ldots 3030304 \ &+ \ldots4571378\ &\overline{\qquad \qquad \qquad }\ & \ \ \ \ldots 7601682 \end{array} \] And \[ \begin{array}{cc}& \ \ \ \ldots 3030304 \ &\times \ldots4571378\ &\overline{\qquad \qquad \qquad }\ & \ \ \ \ldots 4242432 \ & \ \ \ \ldots 212128 \ & \ \ \ \ldots 90912 \ & \ \ \ \ldots 0304 \ & \ \ \ \ldots 128 \ & \ \ \ \ldots 20 \ & \ \ \ \ldots 6 \ &\overline{\qquad \qquad \qquad } \ & \ \ \ \ldots 5038912 \end{array}\] a) Suppose that $A$ is a super number. Prove that there exists a super number $B$ such that $A+B=\stackrel{\leftarrow}{0}$ (Note: $\stackrel{\leftarrow}{0}$ means a super number that all of its digits are zero). b) Find all super numbers $A$ for which there exists a super number $B$ such that $A \times B=\stackrel{\leftarrow}{0}1$ (Note: $\stackrel{\leftarrow}{0}1$ means the super number $\ldots 00001$). c) Is this true that if $A \times B= \stackrel{\leftarrow}{0}$, then $A=\stackrel{\leftarrow}{0}$ or $B=\stackrel{\leftarrow}{0}$? Justify your answer. The code got off so that I can't read what the formula means... Fixed ~dj