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.
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.
There are 39 palindromes from 11 through 399, inclusive.
6*6*3 = 108 numbers.
There are 6 significant digits found in 765.000.
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.
There are 90 four-digit 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
6.
6 digits
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
Some 6 letter palindromes you may like:RedderHannah
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.
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