9 x 10 x 10 x 10 = 9000
Nine is for the 1st and 7th digits. It can be anything from 1-9, just not 0 so you have 9 options. For the 2nd and 6th digits, you can have anything from 0-9 so 10 options. same for 3rd and 5th, and 4th. You don't need to do it the other way around because you'll be repeating the same numbers and you don't want that. Multiply them all together to get 9000. This is data management stuff.
There are 900 6-digit palindromes.
90000. With 10 digit palindromes, the last 5 digits are the same as the first 5 digits in reverse, eg 12345 54321. So it comes down to how many 5 digit numbers are there? They are the numbers "10000" to "99999", a total of 99999 - 10000 + 1 = 90000.
7 digits
There are 400. Assuming the number must be at least 10,000, then: In a 5 digit palindrome, the first and last digits must be the same, and the second and fourth digits must be the same; and: For the first and last digit there is a choice of 4 digits {2, 4, 6, 8}; For each of these there is a choice of 10 digits {0, 1, ..., 9} for the second and fourth digits; For each of the above choices these is a choice of 10 digits {0, 1, ..., 9} for the third digit; Making 4 x 10 x 10 = 400 possible even 5 digit palindromes.
15
There are 90 four-digit palindromes
There are 900 6-digit palindromes.
For there to be palindromes, each digit must be replicated. Therefore there are at most three distinct digits.If there are 3 pairs of different digits, then there are 6 palindromes. If there can be more duplicate digits, then there are 27 palindromes.
Nine. The sum of the digits must be a multiple of 9; because of the repeated digits, this is only possible if the first two digits add up to 9.
81
90000. With 10 digit palindromes, the last 5 digits are the same as the first 5 digits in reverse, eg 12345 54321. So it comes down to how many 5 digit numbers are there? They are the numbers "10000" to "99999", a total of 99999 - 10000 + 1 = 90000.
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, 212, 222
7 digits
Six. 1551 2442 3333 4224 5115 6006
7 digits.