Question

show how the sum of a 4 digit number and the number formed by reversing the order of its digit is divisible by 11​

Answer 1

Answer:

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

Step-by-step explanation:

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.