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×m ends in 1.
Show that, given any n, we can find m such that n×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×m contains all the digits 0,1,2,…,9.