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0

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.

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Related Questions

How many palindromes are between 0 and 1000?

109


How many palindromes are there between 0 and 1?

none you weirdo


How many palindromes are in 0 to10000000?

1999


How many 7's between 0 and 1000?

There are 300 of them.


How many palindromes are there between 10000-99999?

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).


How many palindromes between 900 and 1900?

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.


How many palindromes from 900-1900 and how?

2800


How many palindromes are from 0 to 1000?

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.


How many years is between 250ad and 50bc?

300 mate, 300 * * * * * Wrong, because there wasa no Year 0. The correct answer is 299.


How many palindromes are there between 900 - 1900?

-1000


What is the halfway point between 0 and 600?

(0+600)/2 = 300


Find the number of integer palindromes between 10000 and 99999 that are divisible by 25?

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.