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 B $1$

Let $n=10a+b$ where $a$ is tens place digit and $b$ is units place digit.

Now, for $a^2+b^3$ to be a 2 digit number, the possible combinations of $a$ and $b$ are $(1:5,4),(1:8,3),(2:9,2),(3:9,1),(4:9,0)$ where $k:n⟹$ all integers from $k$ to $n$.

For $b=4$, the units place digit of $a^2+b^3$ will always be $4+$ units place digit of a non-zero cube whereas units place of $n$ will be $4$. Hence, no solution from this set.

For $b=3$, the units place digit of $a^2+b^3$ will always be $7+$ units place digit of a non-zero cube whereas units place of $n$ will be $3$. Only possibility is $a=6⟹n=63$ and $a^2+b^3=63$.

For $b=2$, the units place digit of $a^2+b^3$ will always be $8+$ units place digit of a non-zero cube whereas units place of $n$ will be $2$. Only possibilities are $a=2⟹n=22$ and $a^2+b^3=12$ or $a=6⟹n=62$ and $a^3+b^2=44$ which are not feasible.

For $b=1$, the units place digit of $a^2+b^3$ will always be $1+$ units place digit of a non-zero cube whereas units place of $n$ will be $1$. Hence, no solution exists for this set for this case.

For $b=0$, the units place digit of $a^2+b^3$ will always be $0+$ units place digit of a non-zero cube whereas units place of $n$ will be $0$. Hence, no solution exists for this set for this case.

Hence, only 1 solution when $n=63$ exists.