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

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 $\displaystyle 94$