We know that $ \gcd(m,n) \cdot \mathop{\text{lcm}}[m,n] = mn $ for all positive integers $ m $ and $ n $. Hence, in this case, the other number is \[\frac{(x + 3) \cdot x(x + 3)}{40} = \frac{x(x + 3)^2}{40}.\] To minimize this number, we minimize $ x $. This expression is not an integer for $ x = $ 1, 2, 3, or 4, but when $ x = 5 $, this expression is $ 5 \cdot 8^2/40 = 8 $. Note that that the greatest common divisor of 8 and 40 is 8, and $ x + 3 = 5 + 3 = 8 $. The least common multiple is 40, and $ x(x + 3) = 5 \cdot (5 + 3) = 40 $, so $ x = 5 $ is a possible value. Therefore, the smallest possible value for the other number is $ \boxed{8} $.