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 39 palindromes from 11 through 399, inclusive.
6*6*3 = 108 numbers.
There are 6 significant digits found in 765.000.
Well, isn't that a happy little question! To find out how many five-digit even palindromes there are, let's break it down. A five-digit number has the form ABCBA, where A, B, and C are digits. Since the number is even, A must be even. So, there are 5 options for A (0, 2, 4, 6, 8), 10 options for B (0-9), and 1 option for C. Multiply these options together and you'll find the total number of five-digit even 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.
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
Well, isn't that a happy little question! To find out how many five-digit even palindromes there are, let's break it down. A five-digit number has the form ABCBA, where A, B, and C are digits. Since the number is even, A must be even. So, there are 5 options for A (0, 2, 4, 6, 8), 10 options for B (0-9), and 1 option for C. Multiply these options together and you'll find the total number of five-digit even 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