In a $\displaystyle 2$-digit number, the digit at the unit's place is $\displaystyle 5$ less than the digit at the ten's place. The product of the digits is $\displaystyle 36$ . Find the number. /DA2AB/$\displaystyle 21$

Let digit at ten's place be x then digit at unit's place \(\displaystyle =\mathbf{x}-\mathbf{5}\) \[x(x-5)=36 \] \(\displaystyle \Rightarrow \mathrm{x}^{2}-5 \mathrm{x}-36=0\) \(\displaystyle (x-9)(x+4)=0\) \(\displaystyle \mathrm{x} \neq-4\) so, \(\displaystyle \mathrm{x}=9\) ∴ Required number is $\displaystyle 94$