Question

Prove that a $4-digit$ palindrome is always divisible by $11$

Answer 1

A palindromic number is a number that remains the same when its digits are reversed. For example $9,11,33,101,141,676,1001,1771$ etc. are palindrome numbers. The palindrome number with four digits takes the form $abba$.

Here, the first digit can be chosen in $9$ ways $(1,2,3,4,5,6,7,8,9)$ and second in $10$ ways $(0,1,2,3,4,5,6,7,8,9)$ and the other two digits can be chosen on the basis of first two digits.

So total number of four digit palindromic numbers are $90$ and these are given as:

$1001,1111,1221,1331,1441,1551,1661,1771,1881,1991,.........,9009,$ $9119,9229,9339,9449,9559,9669,9779,$$9889,9999$

These numbers can be written as:

$1001=990+11=11(90+1)$

$1111=1100+11=11(100+1)$

$1221=1210+11=11(110+1)$

and so on.....

Hence, each $4$ digit palindrome number is divisible by $11$.