Question

If the square of a number of two digits is decreased by the square by the square of the number formed by reversing the digits, then the result is not always divisible by:

• A. 9
• B. the product of the digits
• C. the sum of the digits
• D. the difference of the digits
• E. 11

Answer 1

The correct option is B the product of the digits

Let the number be $10a+b$

Number formed by reversing the digits $b=10b+a$

Difference = ${(10a+b)}^2-{(10b+a)}^2=99(a^2-b^2)$

$\therefore$ Result = $99(a+b)(a-b)$

Result is divisible by $9,11$ sum of digits and difference of digits.

$\therefore$ Result is always divisible by product of digits.