11, 22, 33, 44, 55, 66, 77, 88, 99
That's 9
101, 111, 121, 131, 141, 151, 161, 171, 181, 191
That's 10
202, 212, 222, 232, 242, 252, 262, 272, 282, 292
That's 10
Total of 29
Unless 0 - 9 are considered palindromes also.
109
none you weirdo
1999
There are 300 of them.
There are 9,000 palindromes between 10,000 and 99,999. A five-digit palindrome takes the form (abcba), where (a) can range from 1 to 9 (9 options), and (b) and (c) can each range from 0 to 9 (10 options each). Thus, the total number of five-digit palindromes is (9 \times 10 \times 10 = 9000).
109
none you weirdo
1999
There are 300 of them.
There are 9,000 palindromes between 10,000 and 99,999. A five-digit palindrome takes the form (abcba), where (a) can range from 1 to 9 (9 options), and (b) and (c) can each range from 0 to 9 (10 options each). Thus, the total number of five-digit palindromes is (9 \times 10 \times 10 = 9000).
Between 900 and 1900, the palindromes are numbers that read the same forwards and backwards. These are specifically the three-digit numbers in the form of "aba," where "a" is a digit from 9 to 1, and "b" is any digit from 0 to 9. The palindromes in this range are: 909, 919, 929, 939, 949, 959, 969, 979, 989, 999, 1001, 1111, 1221, 1331, 1441, 1551, 1661, 1771, 1881, and 1991. In total, there are 19 palindromes between 900 and 1900.
2800
There are 199 palindromic numbers between 0 and 1000. These include single-digit numbers (0 to 9), two-digit numbers (e.g., 11, 22, ... 99), and three-digit numbers (e.g., 101, 111, ... 999). Each of these categories contributes to the total, with the three-digit palindromes being in the form of ABA, where A and B are digits.
300 mate, 300 * * * * * Wrong, because there wasa no Year 0. The correct answer is 299.
-1000
(0+600)/2 = 300
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.