76(1010). A collective farm team must harvest a potato crop by a certain deadline. After 60% of the potatoes were harvested, a combine was sent to assist the team, which reduced the harvesting time by 5 days. How many days would it have taken to harvest the potatoes without the combine's help, if it is known that the combine could have completed the entire job 8 days faster than the team?
Let's solve it. Let the brigade need $t$ days to harvest potatoes without the help of a combine, then the combine needs $(t-8)$ days to complete the entire work.

In one day, the brigade can complete $\frac{1}{t}$ of the entire work, and the combine can complete $\frac{1}{t-8}$ of the entire work in one day. With the help of the combine, the brigade completes $\frac{2 t-8}{t(t-8)}$ of the work in one day $\left(\frac{1}{t}+\frac{1}{t-8}=\right.$ $\left.=\frac{2 t-8}{t(t-8)}\right) \cdot 60\% \left(\frac{3}{5}\right)$ of the potatoes are harvested by the brigade in $\frac{3 t}{5}$ days $\left(\frac{3}{5} : \frac{1}{t}=\frac{3 t}{5}\right)$, and the remaining $40\% \left(\frac{2}{5}\right)$ of the potatoes are harvested by the brigade with the help of the combine in $\frac{t(t-8)}{5(t-4)}$ days $\left(\frac{2}{5} : \frac{2 t-8}{t(t-8)}=\right.$ $\left.=\frac{t(t-8)}{5(t-4)}\right)$. According to the condition, we have:

$$
t-5=\frac{3 t}{5}+\frac{t(t-8)}{5(t-4)}
$$

from which

$$
\begin{gathered}
t^{2}-25 t+100=0 \\
t_{1}=20, t_{2}=5
\end{gathered}
$$

The condition of the problem ( $t>5$ ) is satisfied by one root $t=20$. Answer: 20 days.

$\square$ 77(1011). Two riders set out simultaneously from $A$ and $B$ towards each other, and one arrives at $B$ after 27 minutes, while the other arrives at $A$ after 12 minutes after the meeting. How many minutes did each rider take to travel the path $A B$?

Solution. Let $A B=s$ (in meters), and the time until the riders meet is $t$ (in minutes). Then the solution of the problem reduces to solving one of the three equations (see the solution of problem 28, p. 112):

$$
\begin{aligned}
& \frac{s}{t+27} \cdot 27=\frac{s}{t+12} \cdot t \\
& \frac{s}{t+27} \cdot t=\frac{s}{t+12} \cdot 12 \\
& \frac{s}{t+27} \cdot t+\frac{s}{t+12} \cdot t=s
\end{aligned}
$$

Answer: the first rider traveled the path $A B$ in 45 minutes, the second in 30 minutes.

$\triangle$ 78. Simplify the expression

$$
70 \cdot\left(71^{9}+71^{8}+71^{7}+\ldots+71^{2}+72\right)+1
$$

Solution. Let's denote the number 71 for brevity as $a$. Then

$$
\begin{gathered}
(a-1)\left(a^{9}+a^{8}+a^{7}+\ldots+a^{2}+a+1\right)+1= \\
=\left(a^{10}+a^{9}+a^{8}+\ldots+a^{3}+a^{2}+a\right)-\left(a^{9}+a^{8}+a^{7}+\ldots+\right. \\
\left.+a^{2}+a+1\right)+1=a^{10}
\end{gathered}
$$

Therefore, the given expression equals $71^{10}$.