656
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.
1001 is the smallest 4 digit palindromic number.
Well, honey, there are 18 palindromic numbers between 10 and 200. Palindromic numbers are like your favorite guilty pleasure - they read the same forwards and backwards. So next time you see a palindromic number, give it a little wink and a nod for being symmetrical in all the right places.
The last digit is always the estimated digit in a number
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)
137?
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.
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No.
1001 is the smallest 4 digit palindromic number.
It can have up to 5 digits.
There are more 12-digit palindromic numbers than 11-digit palindromic numbers. This is because the number of possible 12-digit palindromic numbers is greater than the number of possible 11-digit palindromic numbers. In general, the number of palindromic numbers of length n is 9 * 10^((n-1)/2), so for 11-digit palindromic numbers, there are 9 * 10^5 = 900,000 possibilities, while for 12-digit palindromic numbers, there are 9 * 10^6 = 9,000,000 possibilities.
Let the first and third digits be equal. Choose any other number as the second digit. 202, 919, etc.
17 \
There are 10 digits, but for a three digit number the first number cannot be a 0. Thus: there is a choice of 9 digits for the first (and last digit which must be the same), with 10 choices of digit for the second (middle) digit, making 9 × 10 = 90 such palindromic numbers.
It works!