Question

How many 5-digit palindromes contain only even digits?

Answer 1

To find the number of 5-digit palindromes with only even digits, multiply the number of choices for each digit position. There are 4 choices for the first digit, 4 choices for the second digit, and 5 choices for the third digit, leading to 80 possible palindromes.

Step-by-step explanation:

The question asks how many 5-digit palindromes contain only even digits. A palindrome is a number that reads the same forwards and backwards. Since we are dealing with 5-digit numbers, the first and last digits must be the same, the second and fourth digits must be the same, and the middle digit can be any even digit.

For the first digit, we have 4 choices: 2, 4, 6, or 8 (0 cannot be used because then it wouldn't be a 5-digit number). The second digit also has 4 choices for the same reasons. The middle digit has 5 choices because 0 can now be included. The third and fifth digits will be determined by the first and second digits.

Therefore, we can calculate the total number of palindromic numbers by multiplying the choices for each digit: 4 choices (for 1st digit) × 4 choices (for 2nd digit) × 5 choices (for 3rd digit).

4 × 4 × 5 = 80

So, there are 80 possible 5-digit palindromes with only even digits.

Answer 2

The correct answer is - 900

Step-by-step explanation:

A palindrome is a positive integer that is same whether read from left to right or from right to left. For example, 123321 palindromes.

An n-digit palindrome is determined from the first n/2 digits if n is even, and from the first n+1/2 digits if n is odd.

Therefore, if n is even, there are 9×10⁽n⁻²/²⁾ palindromes; and if n is odd, there are 9×10⁽n₋²/²⁾ palindromes. where n is number of digits.

For n=5 , there are 9×10²=900 palindromes