We know two things about the number:
A: If you reverse it and divide by 2, you get the original number
B: If you multiply it by 2 and reverse it, you get the original number again.
Deducing from there,
* From A, the first digit must be even, otherwise the reversed number wouldn't be divisible by 2
* From B, the first digit must be < 5, otherwise the doubled number would have too many digits (a number with leading zeroes would be ill-formed, so we exclude those straight away)
So the first digit must be either 2 or 4.
* From B, the last digit must be either twice the first digit (if the second digit <5), or one more than that (if the second digit >= 5). (We can omit the usual "modulo 10" bit here, because our numbers are smaller than 5)
* If the first digit is 2, then the last digit must be 4 or 5. However, using A, we get a number ending in 2, divided by 2; the result must end in either 6 or 1. So 2 cannot be the first digit.
* If the first digit is 4, the last digit must be either 8 or 9.
Again using A, a number ending in 4 divided by 2 will end in either 2 or 7. So 4 cannot be the first digit either
It would seem that Ron must have used the single-phase Stetson-Milliner method, because
apart from the trivial 0, such a number doesn't exist at all, and Ron totally pulled his answer out of a hat.
(Hope I didn't make any mistakes there; tried to double-check, but it's difficult to proofread one's own logic because one becomes blind to any flaws.)