[i]Super number[/i] 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 [i]super number[/i]. Note that all of positive integers are [i]super numbers[/i], 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 [i]super numbers[/i]. For example:
\[  \begin{array}{cc}&  \ \ \ \ldots 3030304 \\ &+ \ldots4571378\\ &\overline{\qquad \qquad \qquad }\\ & \ \ \ \ldots 7601682 \end{array} \]

And

\[ \begin{array}{cl}&  \ \ \ \ldots 3030304 \\ &\times \ldots4571378\\ &\overline{\qquad \qquad \qquad }\\ & \ \ \ \ldots 4242432 \\  & \ \ \ \ldots 212128 \\ & \ \ \ \ldots 90912 \\ & \ \ \ \ldots 0304 \\ & \ \ \ \ldots 128 \\ & \ \ \ \ldots 20 \\ & \ \ \ \ldots 6 \\ &\overline{\qquad \qquad \qquad } \\ & \ \ \   \ldots 5038912 \end{array}\]

[b]a)[/b] Suppose that $A$ is a [i]super number[/i]. Prove that there exists a [i]super number[/i] $B$ such that $A+B=\stackrel{\leftarrow}{0}$ (Note: $\stackrel{\leftarrow}{0}$ means a super number that all of its digits are zero).

[b]b)[/b] Find all [i]super numbers[/i] $A$ for which there exists a [i]super number[/i] $B$ such that $A \times B=\stackrel{\leftarrow}{0}1$ (Note: $\stackrel{\leftarrow}{0}1$ means the super number $\ldots 00001$).

[b]c)[/b] Is this true that if $A \times B= \stackrel{\leftarrow}{0}$, then $A=\stackrel{\leftarrow}{0}$ or $B=\stackrel{\leftarrow}{0}$? Justify your answer.
### Part (a)
1. Let \( A = \ldots a_3 a_2 a_1 a_0 \) be a super number.
2. We need to find a super number \( B = \ldots b_3 b_2 b_1 b_0 \) such that \( A + B = \stackrel{\leftarrow}{0} \), which means all digits of the sum are zero.
3. For each digit \( a_i \) of \( A \), choose \( b_i \) such that \( a_i + b_i = 0 \) modulo 10. This means \( b_i = 10 - a_i \) if \( a_i \neq 0 \), and \( b_i = 0 \) if \( a_i = 0 \).
4. Therefore, \( B = \ldots (10 - a_3) (10 - a_2) (10 - a_1) (10 - a_0) \).

Thus, for any super number \( A \), there exists a super number \( B \) such that \( A + B = \stackrel{\leftarrow}{0} \).

\(\blacksquare\)

### Part (b)
1. We need to find all super numbers \( A \) for which there exists a super number \( B \) such that \( A \times B = \stackrel{\leftarrow}{0}1 \).
2. Note that \( \stackrel{\leftarrow}{0}1 \) is the super number \( \ldots 00001 \), which is equivalent to the integer 1.
3. Therefore, we need \( A \times B = 1 \).
4. The only super numbers that satisfy this condition are \( A = 1 \) and \( B = 1 \), because 1 is the multiplicative identity.

Thus, the only super number \( A \) for which there exists a super number \( B \) such that \( A \times B = \stackrel{\leftarrow}{0}1 \) is \( A = 1 \).

\(\boxed{1}\)

### Part (c)
1. We need to determine if \( A \times B = \stackrel{\leftarrow}{0} \) implies \( A = \stackrel{\leftarrow}{0} \) or \( B = \stackrel{\leftarrow}{0} \).
2. Note that \( \stackrel{\leftarrow}{0} \) is the super number \( \ldots 00000 \), which is equivalent to the integer 0.
3. If \( A \times B = 0 \), then either \( A = 0 \) or \( B = 0 \) (or both), because 0 is the absorbing element in multiplication.

Thus, if \( A \times B = \stackrel{\leftarrow}{0} \), then \( A = \stackrel{\leftarrow}{0} \) or \( B = \stackrel{\leftarrow}{0} \).

\(\boxed{\text{True}}\)