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
The smallest 4-digit palindrome is 1001. To find if it can be expressed as the sum of two 3-digit palindromes, consider the smallest 3-digit palindromes, which are 101, 111, 121, etc. The combination of 101 and 900 (another 3-digit palindrome) gives 1001, making 1001 the sum of two 3-digit palindromes. Thus, the answer is 1001.
There are 37 palindromes between from 2 to 299 inclusive.
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.
The smallest 4-digit palindrome is 1001. To find if it can be expressed as the sum of two 3-digit palindromes, consider the smallest 3-digit palindromes, which are 101, 111, 121, etc. The combination of 101 and 900 (another 3-digit palindrome) gives 1001, making 1001 the sum of two 3-digit palindromes. Thus, the answer is 1001.
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.
111, 121, 222, 212
2
There are the same number: any twelve digit palindrome is just an eleven digit palindrome with its central digit repeated. In general, if we are considering numbers of n digits: For even values of n, the number of palindromes is 9 x 10n/2 - 1 For odd values of n, the number of palindromes is 9 x 10(n-1)/2
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.