In a certain job, 3 workers worked; the first for 6 days, the second for 3 days, and the third for 8 days. Per head wages were $36 \mathrm{~K}$ for the first, $12 \mathrm{~K}$ for the second, and $24 \mathrm{~K}$ for the third. How many days would it take for each to complete the job alone?
The entire work was paid $72 \mathrm{~K}-\mathrm{t}$. The first worker received half of the total amount for 6 days of work, so $2 \cdot 6=12$ 

days would have been needed to complete the entire work. The second worker received $\frac{1}{6}$ of the total amount for 3 days, and the third worker received $\frac{1}{3}$ of the total amount for 8 days, so the second worker would have completed the entire work in $6 \cdot 3=18$ and the third worker in $3 \cdot 8=24$ days.

Number of solutions: 29.