jmc

We know that $\gcd (m,n)\cdot \mathrm{lcm}[m,n]=mn$ 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 $\boxed{8}$.