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...
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
There are 9 palindromic numbers between 400 and 500. A palindromic number reads the same forwards and backwards. The palindromic numbers in this range are 404, 414, 424, 434, 444, 454, 464, 474, and 484.
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
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.
1089
10
1089
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.
19
1089 palindromes
1001 - 1991, etc = 10, x (2,3,4,5,6,7,8,9) 8 = 80
410
410