Question

The square of the sum of the digits of a two-digit number is equal to $40$ more than the sum of the squares of its digits. Find the number

• A. $18$
• B. $27$
• C. $36$
• D. Cannot be determined

Answer 1

The correct option is D Cannot be determined

Let the tens' digit be $x$ and the units digit be $y$.
As per the given statement, we have
$(x+y{)}^2=40+x^2+y^2$
$\Rightarrow x^2+y^2+2xy=40+x^2+y^2$
$\Rightarrow 2xy=40$
$\Rightarrow xy=20$
So, $x$ or $y$ can be $5$ or $4$
The numbers can be $54$ or $45$.

Hence, the number cannot be determined.