9 of them.
There are two possible digits for the first and last digit, and two possible digits for the centre digit, making 2 × 2 = 4 possible 3 digit palindromes from the set {1, 2}, namely the set {111, 121, 212, 222}.
111, 121, 222, 212
111, 121, 212, 222
There are 37 palindromes between from 2 to 299 inclusive.
111, 121, 222, 212
There are two possible digits for the first and last digit, and two possible digits for the centre digit, making 2 × 2 = 4 possible 3 digit palindromes from the set {1, 2}, namely the set {111, 121, 212, 222}.
111, 121, 222, 212
111, 121, 212, 222
There are 6 solutions in all.The middle digit must be 0 , 2 or 4.If the middle digit is 0, then the first two digits must sum to 3. There are three possible solutions: 12, 21 and 30.If the middle digit is 2, then the first two digits must sum to 2. There are two possible solutions: 11 and 20.If the middle digit is 4, then the first two digits must sum to 1. There is one possible solutions: 10.
If n is even then number of palindromes is 2^(n/2). If n is odd then number of palindromes is 2*2^[(n-1)/2].
There are 37 palindromes between from 2 to 299 inclusive.
2
111, 121, 222, 212
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.
A palindrome reads the same forwards as backwards. Therefore the first and last digits must be the same, the second and penultimate the same, and so on. As 1000 & 3000 both have 4 digits, there are two pairs of digits that must be the same: the first and fourth, the second and third. The first digit (and hence last digit) can only be 1 or 2 - a choice of 2 The second digit (and hence third digit) can be any of 0, 1, 2, ..., 9 - a choice of 10 for each of the choices of the first/last digit. Thus there are 2 × 10 = 20 palindromes between 1000 and 3000.
Without repeats there are 4 × 3 = 12 possible 2 digit numbers. With repeats there are 4 × 4 = 16 possible 2 digit numbers.
Highest 2-digit number = 99 Highest 1-digit number = 9 Highest possible sum from 2, 2-digit numbers = 198 Highest possible sum from 2, 1-digit numbers = 18