Workload-4

Since the work takes 8 men 13 days to complete, the work $W = 8 \times 13 = 104$ man-days. The actual work done by the increasing number of men in $n$ days is given by:

$\begin{aligned} W_n & = \underbrace{8n + 4(n-1)+4(n-2)+...+4(3)+4(2)+4(1)}_{n \text{ terms}} \\ & = 8n+4(n-1)n - 4\sum_{k=1}^{n-1} k \\ & = 4n^2 + 4n - 4\times \frac {(n-1)n}2 \\ & = 4n^2 + 4n - 2n^2 + 2n \\ & = 2n^2 + 6n \end{aligned}$

For the work to end on the $n$ th day, $W_n \ge W$ . That is $2n^2 + 6n \ge 104 \implies n^2 + 3n \ge 52$ . We note that $n=\boxed{6}$ is the solution.

The work takes 8 × 13 = 104 man-days (which means, that e.g. it would take 104 days for 1 man to finish it or would require 104 men to complete the work in one day.)

$\text {Let the required number of days be } k \in \mathbb {N} \ !$

Now, we can set up the following inequality:

8 + 12 + ... + (8 + 4 × (k - 1)) ≥ 104

After applying the formula regarding the sum of the arithmetic sequence:

$k × \frac {8 + (8 + 4 × (k - 1))}{2} ≥ 104$

$k(8 + 2(k - 1)) ≥ 104$

$2k^2 + 6k - 104 ≥ 0$

$k^2 + 3k - 52 ≥ 0$

After solving this quadratic inequality for k > 0, we get:

$k ≥ \frac {-3 + \sqrt {217}}{2}$

$k ≥ 5.8655$

Therefore, our answer should be:

$\boxed { \text {6th day }}$