Let’s begin by seeing what happens with four digits, five digits, etc.

With four digits there are two numbers that work: 5252 and 5772.

Since in each case abcd/dcba is equivalent to 52/25, abcd must be a multiple of 52 and dcba must be a multiple of 25. The same multiple in fact.

Let’s take a look at exactly what that multiple is for the first few:

52/25 = 1

572/275 = 11

5252/2525 = 101

5772/2775 = 111

52052/25025 = 1001

57772/27775 = 1111

You might notice that the length of each of these factors is always one less than the number of digits of the final number, because multiplying by 52 or 25 increases the digits by 1. You might also notice that each number is palindromic, because the denominator needs to be the reverse of the numerator. And that it only uses digits 1 and 0, because if you had any digits greater than 1, there would be carrying, which would disrupt the digits of the numerator and the denominator in different ways.

For the 25-digit numbers of the question, we need to find all palindromic 24-digit numbers that are made up of 1s and 0s.

The first digit needs to be a 1 (otherwise the number would not be 25 digits long). The next 11 digits can be either 0 or 1, and the remaining 12 digits are defined by the fact that it needs to be a palindrome.

So we have 11 binary decisions to make. 2^11 = 2048. Therefore there are 2048 25-digit numbers which, when divided by their reversal, equal 2.08. You’ll forgive me for not listing them all, but I’ll pick one at random by flipping a coin:

100010110010010011010001:

5200525720520520572520052/2500252750250250275250025 = 2.08