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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: \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 \(0\)