Question

The sum of a two-digit number and the number obtained by reversing the digits is always divisible by

• A. 3
• B. 4
• C. 9
• D. 11

Answer 1

The correct option is D

11

Consider the number ab

$ab=10\times a+b$

Reverse of ab is ba.

$ba=10\times b+a$

Sum of the number and its reverse is,

$ab+ba=10\times a+b+10\times b+a$

$\Rightarrow$ $10\times (a+b)+(a+b)=11\times (a+b)$

It is always divisible by $11$.