is it even reasonable....

The number is divisible by 5 and is 8 digits long so it starts with 5 and ends with 5. The number also is palindromic so we just need to find largest 4 digit number starting with 5 which is divisible by 9. Let x be the 4 digit number.

${x < 6000} \Rightarrow {\frac{x}{9} \leq 666} \Rightarrow {\frac{x}{9} = 666} \Rightarrow {x = 5994}$

Therefore the solution is $\boxed{59944995}$ .

As we have to find the largest number so let the number be 99999999... but as it must be divisible by 45 the last digit must be either 5 or 0 but if we use 0 the the number would become 09999990 which is 7 digit . So we must use 5 . Now we have the number as 59999995 .Now 45= 5 x 9. So, the number must also be divisible by 9. as there are two 5's we need two 4's to make it divisible by 9. So, the three possible combinations are 54999945 , 59499495, 59944995 . Largest among this is 59944995

$45 = 3^2 \times 5$ . Firstly, the number must be divisible by 5. The last digit must be either 0 or 5. But if the last digit is 0, the first has to be 0 too since it is a palindrome but then the number becomes 7-digit. So first and last digit have to be 5.

For the number to be divisible by $3^2 = 9$ , sum of digits must be multiple of 9. For the number to be as big as possible, include as many 9's as possible. But we cannot have all 6 remaining numbers to be 9's as the sum of digits would not be divisible by 9 due to the 2 fives. We have to leave out at least 2-digits in that case so we have $599aa995$ as our number.

Notice that the 9's are placed immediately after the five to maximise the value of the number. For the sum of digits to be divisible by 9, $2(5+a+18)$ must be a multiple of 9 which means $(5+a+18)$ is a multiple of 9 since 2 and 9 are coprime. Simplifying this, we realise $(5+a)$ has to be divisible by 9 since 18 is already a multiple of 9. The only possibility is $5+a = 9$ so that $a=4$ . The desired number is $\boxed{59944995}$