Problem source

In this question we consider only positive, non-zero integers written
out in the usual (decimal) way. We say, for example, that 207 ends
in 7 and that 5310 ends in 1 followed by 0. Show that, if $n$ does
not end in 5 or an even number, then there exists $m$ such that $n\times m$
ends in 1. 

Show that, given any $n$, we can find $m$ such that $n\times m$
ends either in 1 or in 1 followed by one or more zeros. 

Show that, given any $n$ which ends in 1 or in 1 followed by one
or more zeros, we can find $m$ such that $n\times m$ contains all
the digits $0,1,2,\ldots,9$.

Solution source

\begin{array}{c|c}
\text{ends in} & \text{multiply by} \\ \hline 
1 & 1 \\
3 & 7 \\
7 & 3 \\
9 & 9
\end{array}
If if $n = 2^a \cdot 5^b \cdot c$ where $c$ has no factors of $2$ and $5$ then we can multiply by $2^b \cdot 5^a$ to obtain $c$ followed by $0$s. Since $c$ is neither even, nor a multiple of $5$, by the earlier part of the question we can find a multiple such that it ends in $1$. Suppose it is a $k$ digit number, the consider

Now consider $1\underbrace{00\cdots0}_{k\text{ digits}}2\underbrace{00\cdots0}_{k\text{ digits}}\cdots 8\underbrace{00\cdots0}_{k\text{ digits}}9\cdot 0$, then clearly each section will end in the leading digit (ie all digits from $1$ to $9$) and also end with a $0$