Question

The number of all $2$-digit numbers $n$ such that $n$ is equal to the sum of the square of digit in its tens place and the cube of the digit in units place is

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is C $2$

Let tens and units place digits are $a$ and $b$ respectively.
$n=ab10a+b=a^2+b^3\Rightarrow 10a-a^2=b^3-ba(10-a)=b(b+1)(b-1).........\left(1\right)$
As the right hand side is a product of three conscutive number so it is divisible by $6$.
So, the left hand side should be divisible by $6$
$\Rightarrow a(10-a)$ is divisible by $6$
$\Rightarrow a=4,6$
Putting the value of $a$ in $\left(1\right)$ we have
$b=3$
So two numbers are possible $43,63$