A job was completed by the $C$ and $D$ brigades working together. In how many days did they complete it, given that the $C$ brigade alone would have finished the job $c$ days later, and it would have taken $D$ $d$ times longer to complete it alone? Apply the result to the cases $c=10, d=2$ and $c=8, d=3$.

I. solution. Let the total completion time be $n$ days. Thus, $C$ alone would have worked for $n+c$ days, completing $1/(n+c)$ of the work each day, so after $n$ days, $c/(n+c)$ of the work would remain for them. This part was actually completed by $D$, which, according to the other data, is $1/d$ of the work. Thus,

$$
\frac{c}{n+c}=\frac{1}{d}, \quad \text { from which } \quad n=c d-c=c(d-1) \text {. }
$$

Since clearly $d>1$ and $c>0$, we always get a positive result for $n$.

From the first data pair, $n=10$ days, and from the second, $n=16$ days. Indeed, in the first case, $d=2$ means they worked with equal efficiency, so $C$ would have worked for as long as it actually took to complete, $n=c$. However, with $d=3$, $C$ completed twice as much as $D$, so the $c=8$ days are $1/3$ of their solo work time, which is 24 days, and $n$ is $2/3$ of this.

$$
\text { Erzsébet Zentai (Tatabánya, Árpád g. II. o. t.) }
$$

II. solution. With the above notation, in one day $C$ completes $1/(n+c)$ of the work, $D$ completes $1/dn$ of the work, and together they complete $1/n$ of the work.

Thus,

$$
\frac{1}{n+c}+\frac{1}{d n}=\frac{1}{n}, \quad \text { from which } \quad n=c(d-1) \text {. }
$$

Balázs Szabady (Győr, Révai M. g. I. o. t.)