When we multiply a certain two-digit number by the sum of its digits, 567 is achieved. If you multiply the number written in reverse order of the same digits by the sum of the digits, we get 324. Find the number.

The correct option is D: 63

Let the digit at the one's place of a two-digit number be ${}^{\prime }{y}^{\prime }$

and, the digit at ten's place be ${}^{\prime }{x}^{\prime }$

Then the original number $=10x+y$

When the digits are reversed, then the reversed number $=10y+x$

According to the question, we have
$(10x+y)\times (x+y)=567\dots \dots \left(i\right)$ [Given: Original number $\times$ Sum of the digits $=567$]
$(10y+x)\times (x+y)=324\dots \dots \left(ii\right)$ [Given: Reversed number $\times$ Sum of the digits $=324$]

On dividing eq.(i) by (ii), we get

$\frac{10x+y}{10y+x}=\frac{567}{324}$

$\frac{10x+y}{10y+x}=\frac{567÷81}{324÷81}$ [HCF of 567 and 324 is 81]

$\frac{10x+y}{10y+x}=\frac{7}{4}$

$\Rightarrow (10x+y)\times 4=(10y+x)\times 7$

$\Rightarrow 40x+4y=70y+7x$

$\Rightarrow 40x-7x=70y-4y$

$\Rightarrow 33x=66y$

$\Rightarrow x=2y$

Putting the above value of x in eq.(i), we get

$(10\times 2y+y)\times (2y+y)=567$

$\Rightarrow 21y\times 3y=567$

$\Rightarrow y^2=\frac{567}{63}=9$

$\Rightarrow y=3$

So, $x=2y=2\times 3=6$

Hence, the original number is $63.$