13.373 A computing machine was given the task to solve several problems sequentially. Registering the time spent on the assignment, it was noticed that the machine spent the same multiple of time less on solving each subsequent problem compared to the previous one. How many problems were proposed and how much time did the machine spend solving all the problems, if the time spent on solving all the problems except the first one was 63.5 minutes, the time spent on solving all the problems except the last one was 127 minutes, and the time spent on solving all the problems except the first two and the last two was 30 minutes?
Solution. Let $n$ be the number of problems, $b$ be the time to solve the first problem, and $q$ be the common ratio of the geometric progression that represents the times to complete the problems. Using the formula for the sum of $k$ terms of a geometric progression, we have:

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ b q \cdot \frac { q ^ { n - 1 } - 1 } { q - 1 } = \frac { 127 } { 2 } , } \\
{ b \cdot \frac { q ^ { n - 1 } - 1 } { q - 1 } = 127 , } \\
{ b q ^ { 2 } \cdot \frac { q ^ { n - 4 } - 1 } { q - 1 } = 30 }
\end{array} \Rightarrow \left\{\begin{array} { l } 
{ q = \frac { 1 } { 2 } , } \\
{ b \cdot \frac { q ^ { n - 1 } - 1 } { q - 1 } = 127 , } \\
{ b \cdot \frac { q ^ { n - 2 } - q ^ { 2 } } { q - 1 } = 30 }
\end{array} \Rightarrow \left\{\begin{array}{l}
q=\frac{1}{2} \\
\frac{q^{n-1}-1}{q^{n-2}-q^{2}}=\frac{127}{30}
\end{array} \Rightarrow\right.\right.\right. \\
& \Rightarrow\left\{\begin{array}{l}
q=\frac{1}{2}, \\
q^{n-2}=\left(\frac{1}{2}\right)^{6}
\end{array} \Rightarrow n=8, \Rightarrow b=64, \Rightarrow 64+63.5=127.5\right. \text { (total time to solve all problems). }
\end{aligned}
$$

Answer: 8 problems; 127.5 minutes.