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 $$\frac{(x+5)\cdot x(x+5)}{50}=\frac{x(x+5{)}^2}{50}.$$To minimize this number, we minimize $x$.

We are told that the greatest common divisor is $x+5$, so $x+5$ divides 50. The divisors of 50 are 1, 2, 5, 10, 25, and 50. Since $x$ is a positive integer, the smallest possible value of $x$ is 5. When $x=5$, the other number is $5\cdot 10^2/50=10$.

Note that that the greatest common divisor of 10 and 50 is 10, and $x+5=5+5=10$. The least common multiple is 50, and $x(x+5)=5\cdot (5+5)=50$, so $x=5$ is a possible value. Therefore, the smallest possible value for the other number is $\boxed{10}$.