109
none you weirdo
1999
There are 300 of them.
300 mate, 300 * * * * * Wrong, because there wasa no Year 0. The correct answer is 299.
-1000
(0+600)/2 = 300
30 times because when you count up in tens this is the answer
Numbers that end in 0 or 5 are all divisble by 5
All palindromes between 10000 and 99999 are of the form abcba where a is a digit in {1, 2, ..., 9} and b and c are digits in {0, 1, ..., 9} All multiples of 25 end in 00, 25, 50 or 75. As a cannot be 0, this means the only multiple of 25 we are interested in are those which end in 25 and 75. This means we are looking for integers of the form 52c25 and 57c75. There are 10 possible values for the digit c in each case (one of {0, 1, ..., 9}), which means: There are 20 palindromes between 10,000 and 99,999 which are divisible by 25.
49 zeros [leading zeros are not used like 001, 002, etc.] This does not include the two zeros in 300 (since the question stated between). So if 300 is included, then 51 zero digits.
0