46(1122). Several points are marked on a plane, no three of which lie on the same line. A line is drawn through every two points. How many points are marked on the plane if it is known that a total of 45 lines have been drawn?
Solution. The problem reduces to solving the equation $\frac{x(x-1)}{2}=45$ in natural numbers, where $x$ is the number of points marked on the plane. We have $x^{2}-x-90=0$, from which $x_{1}=10, x_{2}=-9$. Only the root $x=10$ satisfies the condition of the problem.

Answer: 10 points.

$\triangle 47(969)$ '. Prove that to find the square of a two-digit number ending in the digit 5 and having $n$ tens, it is sufficient to multiply the number of tens $n$ by $n+1$ and append 25 (see problem 1091 from [4]).

Solution. Let's square the two-digit number $a 5=10 a+5$. We have:

$$
(10 a+5)^{2}=100 a^{2}+100 a+25=100 a(a+1)+25
$$

We obtained that the square of the number $\overline{a 5}$ ends with the number 25, and the number of its hundreds is $a(a+1)$, which is what we needed to prove.

Remark. It is also advisable to solve problem 856 from $[4]$.

' For problems 47-77, the numbers in parentheses indicate the exercise numbers in the textbook [2] 122
$\triangle$ 48(970). Find a two-digit number whose square is written using the digits $0 ; 2 ; 3$ and 5.

$\mathrm{Pe}$ solution. Since the square of an integer cannot end in the digit 2, 3, or a single zero, the square of an integer can only end in the digit 5 from the four given digits. Then the second-to-last digit is 2. Considering that the first digit cannot be zero, we establish that the square of a two-digit number is 3025, and the sought number is 55. (To find the sought number, it is useful to use the previous problem.)