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well, i think if you use this you can find out. A = 1-9 ,B = 0-9 , C = 0-9 , D = 0-9 , E = 0-9 for 2digit numbers = A A for 3 digit numbers = A B A for 4 digit numbers = A B B A and so on till you get to for 8 digit numbers = A B C D D C B A for 9 digit numbers = A B C D E D C B A and last for 10 digit number = A B C D E E D C B A this should work...

Q: How many 10 digit palindromic numbers are there?

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To be a 3 digit palindromic number, it must be of the form aba.I assume that a 3 digit number must be at least 100 (so that 020 for example does not count as a 3 digit number):a can be any of the nine digits 1-9;for each of these b can be any of the ten digits 0-9Thus there are 9 x 10 = 90 three digit palindromic numbers.

19

there are 10 palindromic numbers between 9000 and 10000 9009,9119,9229,9339,9449,9559,9669,9779,9889,9999

Number of 7 digit combinations out of the 10 one-digit numbers = 120.

There are 5! = 120 such numbers.

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there are 10 palindromic numbers between 9000 and 10000 9009,9119,9229,9339,9449,9559,9669,9779,9889,9999!!!

No.

To be a 3 digit palindromic number, it must be of the form aba.I assume that a 3 digit number must be at least 100 (so that 020 for example does not count as a 3 digit number):a can be any of the nine digits 1-9;for each of these b can be any of the ten digits 0-9Thus there are 9 x 10 = 90 three digit palindromic numbers.

If you think about the digits, you can rewrite them as ABBA, with A being one digit and B being another: A can be 1-9 and B can be 0-9. Since A has to be 1, B can be 0-9, leaving 10 palindromic numbers.

1089

10

1089

19

1089 palindromes

1001 - 1991, etc = 10, x (2,3,4,5,6,7,8,9) 8 = 80

410

410