77(1198). Find the coefficients $a, b$ and $c$ of the polynomial $a x^{2}+b x+c$, knowing that the equality

$$
2 x^{4}+x^{3}-41 x^{2}+83 x-45=\left(a x^{2}+b x+c\right)\left(x^{2}+4 x+9\right)
$$

is an identity.
Solution. By equating the coefficients of the fourth and third powers of $x$ on the left and right sides of the given equality, we will have $a=2, 4a+b=1$, from which $b=-7$.

By equating the free terms, we find $c: -45=-9c, c=5$. Answer: $a=2, b=-7, c=5$.

$\triangle 78(1199)$. Form a polynomial of the sixth degree with the variable $x$, knowing that for any values of $x$ its value is: a) positive; b) negative.

Solution. a) The condition of the problem is satisfied, for example, by the polynomial $x^{6}+x^{4}+x^{2}+1$ or the binomial $x^{6}+0.1$.

b) The condition of the problem is satisfied, for example, by the binomial $-x^{6}-1$.

$79(1200)$. Find all pairs of natural numbers that satisfy the equation $11x+8y=104$.

Solution. Express one of the unknowns from the given equation, for example, $x$:

$$
x=\frac{104-8y}{11}=9+\frac{5-8y}{11}
$$

By trial, we find the smallest natural number $y$ for which $x$ is a natural number: $y=2$, then $x=8$. This value of $y$ is the only one, since for $y>2$ ($y$ is a natural number) $x \notin \mathbb{N}$.

Answer: $(8; 2)$.

$\triangle \mathbf{80}(949)'$. Prove that if a three-digit number is subtracted from a three-digit number written with the same digits as the first but in reverse order, then the modulus of the obtained difference will be divisible by 9 and 11.

Solution. Let $\overline{abc}$ be a three-digit number. Then $\overline{abc}-\overline{cba}=100a+10b+c-100c-10b-a=99a-99c=99(a-c)=$ $=9 \cdot 11(a-c)$.

Since the modulus of the obtained difference is divisible by 9 and 11, the statement is proved.

Remark. In the condition of the problem, instead of the modulus of the obtained difference, we can consider the difference of the given numbers, since when $a>c$ we get a natural number divisible by 9 and 11, and when $a<c$ we get an integer negative number, also divisible by 9 and 11.

$\triangle 81(950)$. Prove that the modulus of the difference of the squares of two consecutive odd natural numbers is equal to twice the sum of these numbers.

Solution. Let $2n-1$ and $2n+1$ be two consecutive odd natural numbers. We have:

$$
\begin{gathered}
\left|(2n-1)^{2}-(2n+1)^{2}\right|=|-8n|=8n \\
2((2n-1)+(2n+1))=8n .
\end{gathered}
$$

The statement is proved.

$\triangle(951)$. If the two-digit number $x$ is inserted between its digits, then the resulting four-digit number will be 66 times greater than the original two-digit number. Find $x$.

Solution. Let $x=\overline{ab}=10a+b$. According to the condition, we have:

$$
\overline{aabb}=66 \cdot \overline{ab}
$$

or

$$
1000a+100a+10b+b=66(10a+b)
$$

from which $440a=55b, 8a=b, a=1, b=8, \overline{ab}=18$.

Answer: $x=18$.

Remark. The problem is analogous to problem 1152 from [1], so one of them can be solved in class, and the other can be assigned as homework.