Question

If we add the square of the digit in the tens place of a positive two digit number to the product of the digits of that number, we shall get $52$ and if we add the square of the digit in the units place to the same product of the digits, we shall get $117$. Find the two-digit number.

Answer 1

Let $n(a,b)$ be two digit number

$n(a,b)=10a+b$

$\text{(square of of digits in ten's place)+(product of digits)}=2$

$a^2+ab=52\to \left(1\right)$

$\text{(square of digit in unit place)+(product of digits)}=117$

$b^2+ab=117\to \left(2\right)$

$\left(1\right)+\left(2\right)\Rightarrow (a+b{)}^2=169$

$\therefore a+b=13(\because a\ge 1,b\ge 0)$

$\left(1\right)-\left(2\right)\Rightarrow a^2-b^2=65$

$-a-b=-5$

$a=4,b=9$

$\therefore n(a,b)=49$ is the required number.