5. Find all $a$ for which the system

$$
\left\{\begin{array}{l}
x^{2}+4 y^{2}=1 \\
x+2 y=a
\end{array}\right.
$$

has a unique solution. If necessary, round it to two decimal places. If there are no solutions, put 0 in the answer.
Solution. Since $2 y=a-x$, from the first equation we get:

$$
x^{2}+(a-x)^{2}=1 \Longleftrightarrow 2 x^{2}-2 a x+a^{2}-1=0
$$

This equation has a unique solution when $\frac{D}{4}=2-a^{2}=0$. Therefore, $a= \pm \sqrt{2}$, and the smallest value is $-\sqrt{2} \approx-1.414214 \ldots$

Answer: $-1.41$.

I-A. The sum of 28218 natural numbers is $2016 \cdot 15$, and their product is $-\left(2016^{2}+15\right)$. Find all possible sets of such numbers. In the answer, indicate the sum of the largest and smallest numbers from all the found sets. If such numbers do not exist, then in the answer, indicate the number 0.

Solution. From the factorization $2016^{2}+15=3 \cdot 19 \cdot 113 \cdot 631$, we conclude that in the problem

$$
\left\{\begin{array}{l}
a_{1}+\ldots+a_{28218}=2016 \cdot 15 \\
a_{1} \cdot \ldots \cdot a_{28218}=2016^{2}+15=3 \cdot 19 \cdot 113 \cdot 631
\end{array}\right.
$$

the numbers $a_{k}$ not equal to 1 can be no more than 4 (i.e., no more than the number of prime factors), so the problem is equivalent to the following

$$
\left\{\begin{array}{l}
a_{1}+a_{2}+a_{3}+a_{4}=2026 \\
a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4}=3 \cdot 19 \cdot 113 \cdot 631 \\
a_{5}=a_{6}=\ldots=a_{28218}=1
\end{array}\right.
$$

We solve the last problem: The set $a_{1}, a_{2}, a_{3}, a_{4}$ consists of the numbers $1,19,113,3 \cdot 631$.

It remains to calculate $(3 \cdot 631)+1=1894$.

Answer: 1894.