$36 \cdot 26 \quad 10$ people form a circle, each person chooses an integer and tells it to their two neighbors, then each person calculates and announces the average of the numbers chosen by their two neighbors, as shown in the figure. What is the number chosen by the person who announced $6$?
(A) 1 .
(B) 5 .
(C) 6 .
(D) 10 .
(E) Not uniquely determined.
(41st American High School Mathematics Examination, 1990)
[Solution] Let the number chosen by the person who announced 6 be $x$, then the number chosen by the person who announced 4 is $5 \cdot 2 - x = 10 - x$, the number chosen by the person who announced 2 is $3 \cdot 2 - (10 - x) = x - 4$, the number chosen by the person who announced 10 is $1 \cdot 2 - (x - 4) = 6 - x$, the number chosen by the person who announced 8 is $9 \cdot 2 - (6 - x) = 12 + x$, and thus, the number chosen by the person who announced 6 is $7 \cdot 2 - (12 + x) = 2 - x$. Therefore, $x = 2 - x$, which gives $x = 1$. Hence, the answer is $(A)$.