If we divide a two digit number by the sum if its digits we get 4 as a quotient and 3 as a remainder. Now, if we divide that two digit number by the product of its digits, we get 3 as a quotient and 5 as a remainder. Find the two digit number.

Let the number be $10x + y$. From the first condition, $10x + y = 4(x + y) + 3$, which simplifies to $2x - y = 1$. From the second, $10x + y = 3xy + 5$. Substituting $y = 2x - 1$ into the second equation gives $10x + 2x - 1 = 3x(2x - 1) + 5$, leading to $6x^2 - 15x + 6 = 0$. Solving gives $x=2$, so $y=3$. The number is 23.