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 $$\frac{(x+2)\cdot x(x+2)}{24}=\frac{x(x+2{)}^2}{24}.$$To minimize this number, we minimize $x$.

This expression is not an integer for $x=$ 1, 2, or 3, but when $x=4$, this expression is $4\cdot 6^2/24=6$.

Note that that the greatest common divisor of 6 and 24 is 6, and $x+2=4+2=6$. The least common multiple is 24, and $x(x+2)=4\cdot (4+2)=24$, so $x=4$ is a possible value. Therefore, the smallest possible value for the other number is $\boxed{6}$.