61 (1161). When the polynomial $2 x^{3}-5 x^{2}+7 x-8$ is multiplied by the polynomial $a x^{2}+b x+11$, the resulting polynomial does not contain either $x^{4}$ or $x^{3}$. Find the coefficients $a$ and $b$ and determine what polynomial results from the multiplication.
Solution. Let's multiply the given polynomials, writing their product (as well as the factors) in standard form:

$$
\begin{aligned}
& \left(2 x^{3}-5 x^{2}+7 x-8\right) \cdot\left(a x^{2}+b x+11\right)= \\
= & 2 a x^{5}-5 a x^{4}+7 a x^{3}-8 a x^{2}+2 b x^{4}-5 b x^{3}+ \\
+ & 7 b x^{2}-8 b x+22 x^{3}-55 x^{2}+77 x-88= \\
= & 2 a x^{5}+(2 b-5 a) x^{4}+(7 a-5 b+22) x^{3}+ \\
& +(7 b-8 a-55) x^{2}+(77-8 b) x-88 .
\end{aligned}
$$

66

According to the condition, the coefficients of $x^{4}$ and $x^{3}$ are 0, i.e., we have the system of equations:

$$
\left\{\begin{array}{l}
2 b-5 a=0 \\
7 a-5 b+22=0
\end{array}\right.
$$

from which $a=4, b=10$.

Answer: $a=4, b=10$; the product is $8 x^{5}-17 x^{2}-3 x-88$.

$62(1162)$. A student told his friend that he had thought of a two-digit number, subtracted from it the number written with the same digits but in reverse order, and obtained the square of an even number. After some thought, the friend stated that the obtained difference was 36. Was he right? Find all two-digit numbers that have the property noticed by the student.

Solution. Let the thought number be $\overline{a b}$. Then

$$
\overline{a b}-\overline{b a}=10 a+b-(10 b+a)=9(a-b)
$$

The number $9(a-b)$ can be the square of an even number only if $a-b$ is the square of an even number. But $a-b<10$ (since $a$ and $b$ are digits), and the only square of an even number less than 10 is 4. Therefore, the difference of the two-digit numbers is 36. The friend was right.

From the condition $a-b=4$, we find by trial $a=9, b=5 ; a=8$, $b=4 ; a=7, b=3 ; a=6, b=2 ; a=5, b=1$.

Thus, the thought number could have been one of the following: $95 ; 84 ; 73 ; 62 ; 51$.

$63(1166)$. How many divisors does the number $10^{10}$ have?

Solution. $10^{10}=(2 \cdot 5)^{10}=2^{10} \cdot 5^{10}$. The divisors of this product have the form $2^{k} \cdot 5^{n}$, where $k$ and $n$ are non-negative integers not greater than 10. Among them, there are 11 divisors containing only powers of 2:

$$
2^{0}, 2^{1}, 2^{2}, \ldots, 2^{10}
$$

and as many divisors containing only powers of 5:

$$
5^{0}, 5^{1}, 5^{2}, \ldots, 5^{10}
$$

We will get all the divisors of the given number if we multiply each of the divisors of the first kind (powers of 2) by each of the divisors of the second kind (powers of 5). In total, we have $11 \cdot 11=121$ divisors.