Question

If the square of a $2$-digit number is decreased by the square of the number formed by reversing the digits, then the result is not always divisible by _________.

• A. $9$
• B. Product of the digits
• C. Sum of the digits
• D. Difference of the digits

Answer 1

Answer: (B)

Product of the digits

Let digit at unit's place$=x$
Digit at ten's place$=y$
$\therefore$ Original no.$=x+10y⟶1$
When digits are reversed,
digit at unit's place$=y$ & at ten's place$=x$
$\therefore$ New no.$=y+10x⟶2$
Let $z={(x+10y)}^2-{(y+10x)}^2$
$=x^2+100y^2-20xy-(y^2+100x^2+20xy)$
$=99y^2-99x^2$
$z=99(y-x)(y+x)$
$\therefore$ z is divisible by $9,(x+y)\&(x-y)$
$\therefore$ It is not always divisible by xy i.e., product of digits.