Olimpiadi della Matematica

Solution:

The answer is (E). Adding the first equation to the third, we obtain 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, subtracting from this equation the second equation of the system multiplied by two, we obtain 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 does indeed have a solution (in fact it has infinitely many possible ones!). For instance, by also setting $x_{4}=x_{5}=\ldots=x_{30}=0$, it is possible to compute values of $x_{1}, x_{2}, x_{3}$ that satisfy the system through successive substitutions.