Question

Prove that a 4 digit palindrome is always divisible by 11.

Answer 1

Hint:In this question use the basic definition of a palindrome that is a number which when reversed gives the exact same number. So assume any number of 4 digit palindromic numbers just like 11, 22, 33, 44, 55 etc. Then to them apply the basic criteria for divisibility by 11 as this will help approaching the problem.

Complete step-by-step answer:
Palindrome numbers:
A palindrome is a number in which when the digits of a number is reversed (I.e. in a two digit number second digit is replaced by first digit and first digit is replaced by second digit) the number remains the same.
For example: 11, 22, 33, 44, 55 etc.
So as we see from the above examples when the digit is replaced the number remains the same so these numbers are called palindrome numbers.
Now we have to prove 4 – digit palindrome numbers are always divisible by 11.
Proof –
According to the above property of palindrome numbers in a four digit palindrome, the first and fourth digits should be the same and the second and third digits should be the same so that the reverse of the numbers is the same.
So the four digit palindrome numbers are
1001, 1111, 1221, 1331, etc. (number starting from one should always end with one).
If number starting from 2 so the palindrome numbers are
2002, 2112, 2222, 2332, etc.
So similarly there are lots of 4 digit palindrome numbers
1001, 1111, 1221, ........................., 9119, 9229, 9339,........., 9889, 9999
Therefore,
1001 = 11(91)
1111 = 11(101)
.
.
.
.
9889 = 11(899)
9999 = 11(909)
Hence each 4 digit palindrome number is divisible by 11 as each number can be expressed as multiple of 11.
Hence proved.

Note:This problem can be approached by another method that is we figure out any palindromic number which is of 4 digit and apply basic divisibility by 11 rule that is if the alternating sum of digits is divisible by 11 then the number will be divisible by 11. This can also give us the answer. That is to say 9889 is a 4 digit palindromic number now alternating sum will be 9-8+8-9=0, clearly 0 is divisible by 11 and so 9889. The alternating sum means take the digits from left to right and follow the pattern -,+,-,+…………..