If I multiply $51,249,876$ by $3$ (thus using all the nine digits once, and once only), I get $153,749,628$ (which again contains all the nine digits once). Similarly, if I multiply $16,583,742$ by $9,$ the result is $149,253,678,$ wherein each case all the nine digits are used. Now, take $6$ as your multiplier and try to arrange the remaining eight digits so as to produce by multiplication a number containing all nine once, and once only. You will find it far from easy, but it can be done.
Solutions: 1
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