Question:
The greatest common divisor of two integers is $(x+3)$ and their least common multiple is $x(x+3)$, where $x$ is a positive integer. If one of the integers is 40, what is the smallest possible value of the other one?

Answer:
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 $\boxed{8}$.