The same as the number of two-digit numbers, since the last two digits
must the same as the first two, only reversed.
So I'll say there are 100 four-digit palindromes.
90
9
There are 900000 of them.
The same as the number of two-digit numbers, since the last two digits must the same as the first two, only reversed. So I'll say there are 100 four-digit palindromes.
899100 of them. Only 900 of the 900000 are palindromic.
90
9
There are 900000 of them.
The same as the number of two-digit numbers, since the last two digits must the same as the first two, only reversed. So I'll say there are 100 four-digit palindromes.
Oh, dude, there are 90 three-digit palindromic numbers. You see, a three-digit palindrome has the form "ABA," where A and B can be any digit from 1 to 9. So, you just multiply the possibilities for A and B, which is 9 choices each, and voilà, you get 9 x 10 = 90. Easy peasy, lemon squeezy!
899100 of them. Only 900 of the 900000 are palindromic.
A palindromic number can have as few as one digit, and as many as infinity. The smallest palindromic number is zero and the largest is '999... to infinity'.
There is: 101,111,121,131,141,151,161,171,181,191 202,212,222,etc... 999 There are 90 palindromic 3 digit numbers
40. Excluding numbers starting with 0.
a 8-digits palindromic number is a number consisting of a 4-digits number written and then written backward, i.e. 1234 4321 so there are as many palindromic 8-digits numbers as 4-digits numbers so from 1000 to 9999, so there are 8999 palindromic 8 digits number (I assumed that 00011000 is not to be considered as a valid 8 digit number)
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.
1 to 1000. Pilandromic 108