$36 \cdot 15$ In a certain office, six people $A, B, C, D, E, F$ take turns working the night shift. The order of the shifts is $A \rightarrow B \rightarrow C \rightarrow D \rightarrow E \rightarrow F \rightarrow A \rightarrow B \cdots$. In the first week, Sunday is covered by $A$, Monday by $B$, $\cdots$, Friday by $F$, and Saturday again by $A$, and so on. Therefore, in the 50th week, $A$ will be on duty on
(A) Saturday.
(B) Friday.
(C) Thursday.
(D) Wednesday.
(China Guangzhou, Wuhan, Fuzhou, and other five cities junior high school mathematics league, 1992)
[Solution] The 50th week starts on a Monday, which is $7 \cdot 49 + 1 = 344$ days from the initial shift.

When the number of days $n \div 6$ has a remainder of 0, it is $A$'s turn; a remainder of 1, it is $B$'s turn; a remainder of 2, it is $C$'s turn; .....

Since $344 \div 6 = 57 \cdots 2$, we know that the start of this week (Monday) is $C$'s turn. After $D$, $E$, $F$, it will be $A$'s turn on Friday.
Therefore, the answer is $(B)$.