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Question: Number of \(5\) digits even palindromes is: A) \(400\) B) \(500\) C) \(4000\) D) \(5000\)...

Number of 55 digits even palindromes is:
A) 400400
B) 500500
C) 40004000
D) 50005000

Explanation

Solution

A palindrome is a number that reads the same forward and backward, such as 242242. The five digit palindrome is of the form of ABCBAABCBA; where AA is the same digit on 1st1^{st} and 5th5^{th} place, BB is the same digit on 2nd2^{nd} and 4th4^{th} place, CC is the digit on 3rd3^{rd} place.

Complete step-by-step answer:
Let the five digit palindrome is of the form of ABCBAABCBA; where AA is the same digit on 1st1^{st} and 5th5^{th} place, BB is the same digit on 2nd2^{nd} and 4th4^{th} place, CC is the digit on 3rd3^{rd} place.
The palindrome must be even, if the number in position AA can only be 2,4,62,4,6 or 88. This number cannot be 00 because if the first digit is 00, it would make the 55 digit number a 44 digit number. So, the place AA can be filled by 44 ways.
The number in position BBcould be 0,1,2,3,4,5,6,7,8,90,1,2,3,4,5,6,7,8,9. So the place BB can be filled by 1010 ways.
Also, the number in position CC could be the same, i.e., 0,1,2,3,4,5,6,7,8,90,1,2,3,4,5,6,7,8,9. So the place CC can be filled by 1010 ways.
So, the desired number of 55 digits even palindromes=4×10×10 = 4 \times 10 \times 10 =400 = 400

Option A is the correct answer.

Note: An another method to solve this problem is described as follows:
The first and last digits can only be even integers and are the same in 44 ways, i.e., 2,4,62,4,6 or 88.
The second and fourth digits are the same but ranges from 090 - 9 in 1010 ways.
The third digit could be any integer from 090 - 9 in 1010 ways.
So the desired number of 55 digits even palindromes =4×10×10=400 = 4 \times 10 \times 10 = 400