9. The real numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{30}$ satisfy the following conditions:

$$
\left\{\begin{array}{l}
20^{3} x_{1}+21^{3} x_{2}+22^{3} x_{3}+\cdots+49^{3} x_{30}=13 \\
21^{3} x_{1}+22^{3} x_{2}+23^{3} x_{3}+\cdots+50^{3} x_{30}=1 \\
22^{3} x_{1}+23^{3} x_{2}+24^{3} x_{3}+\cdots+51^{3} x_{30}=19
\end{array}\right.
$$

What is the value of $21 x_{1}+22 x_{2}+23 x_{3}+\cdots+50 x_{30}$?
(A) There are multiple acceptable values
(B) The system has no solution
(C) 1065
(D) 7
(E) 5
9. The answer is (E). By adding the first equation to the third, one obtains an equation of the form $a_{1} x_{1}+a_{2} x_{2}+\ldots+a_{30} x_{30}=32$, where $a_{1}=(21+1)^{3}-(21-1)^{3}=21^{3}+3 \cdot 21^{2}+3 \cdot 21+1+$ $21^{3}-3 \cdot 21^{2}+3 \cdot 21-1=2 \cdot 21^{3}+6 \cdot 21, a_{2}=(22+1)^{3}-(22-1)^{3}=2 \cdot 22^{3}+6 \cdot 22$, and in general $a_{i}=2 \cdot(20+i)^{3}+6 \cdot(20+i)$. Now, by subtracting from this equation the second equation of the system multiplied by two, one obtains the equation $6 \cdot 21 x_{1}+6 \cdot 22 x_{2}+\ldots+6 \cdot 50 x_{30}=$ $32-2=30$. Consequently, if $x_{1}, \ldots, x_{30}$ satisfy the three equations of the system, then $21 x_{1}+22 x_{2}+\ldots+50 x_{30}=30 / 6=5$.

On the other hand, the system actually has a solution (indeed, it has infinitely many possible solutions!). Even by setting, for example, $x_{4}=x_{5}=\ldots=x_{30}=0$, it is possible to calculate values of $x_{1}, x_{2}, x_{3}$ that satisfy the system through successive substitutions.