For a number to be a multiple of $$45$$ it must be a multiple of $$5$$ and also of $$9$$. In order to be a multiple of $$5$$, a number\textquotesingle s units digit must be $$0$$ or $$5$$. However, the units digit of a palindromic number cannot be $$0$$, so it may be deduced that any palindromic number which is a multiple of $$45$$ both starts and ends in the digit $$5$$. In order to make the desired number as large as possible, its second digit should be $$9$$ and for it to be as small as possible its second digit should be $$0$$. So, if possible, the numbers required are of the form \textquotesingle$$59x95$$\textquotesingle{} and \textquotesingle$$50y05$$\textquotesingle{} . In addition, both numbers are to be multiples of $$9$$ which means the sum of the digits of both must be a multiple of $$9$$. For this to be the case, $$x=8$$ and $$y=8$$ , giving digit sums of $$36$$ and $$18$$ respectively. So the two required palindromic numbers are $$59895$$ and $$50805$$. Their difference is $$9090$$. 
A palindromic number is a number that reads the same when the order of its digits is reversed. What is the difference between the largest and smallest five-digit palindromic numbers that are both multiples of $$45$$? 

  - $$9180$$ 
  - $$9090$$ 
  - $$9000$$ 
  - $$8910$$ 
  - $$8190$$ 