Problem 3. We will call a natural number odd-powered if all its prime divisors enter its factorization with an odd exponent. What is the maximum number of odd-powered numbers that can occur consecutively?

## Answer: 7.
Solution. Note that among any eight consecutive natural numbers, there will definitely be a number that is divisible by 4 but not by 8. Then the number 2 will appear in its factorization to the second power.

Example of seven odd-power numbers: $37,38,39,40,41,42,43$.

Problem 4/1. On side $A C$ of triangle $A B C$, a point $D$ is chosen. It is known that $\angle B A C=30^{\circ}, \angle D B C=75^{\circ}, \angle B C A=45^{\circ}$. Find $C D$, if it is known that $B A+A D=14$.

Answer: 7.

![](https://cdn.mathpix.com/cropped/2024_05_06_e1ba744b92c47f397b79g-4.jpg?height=250&width=515&top_left_y=93&top_left_x=469)

Fig. 1: to the solution of problem $4 / 1$

Solution. From the sum of the angles of triangle $A B C$, it follows that $\angle A B D=30^{\circ}$. Therefore, $A D=B D$.

Draw the height $B H$ in triangle $A B C$ (Fig. 1). Triangle $B H C$ is a right isosceles triangle, so $C H=B H$. Triangles $A B H$ and $B D H$ are right triangles with an angle of $30^{\circ}$, so in them the legs opposite the angle of $30^{\circ}$ are half the hypotenuse. It turns out that $2 B H=A B, 2 H D=B D$.

Then

$$
C D=C H+H D=B H+H D=\frac{A B}{2}+\frac{B D}{2}=\frac{A B+A D}{2}=7
$$

Problem 4/2. On side $A C$ of triangle $A B C$, a point $D$ is chosen. It is known that $\angle B A C=30^{\circ}, \angle D B C=75^{\circ}, \angle B C A=45^{\circ}$. Find $C D$, if it is known that $B A+A D=16$.

Answer: 8.

Solution. From the sum of the angles of triangle $A B C$, it follows that $\angle A B D=30^{\circ}$. Therefore, $A D=B D$.

![](https://cdn.mathpix.com/cropped/2024_05_06_e1ba744b92c47f397b79g-4.jpg?height=261&width=507&top_left_y=1327&top_left_x=469)

Fig. 2: to the solution of problem $4 / 2$

Draw the height $B H$ in triangle $A B C$ (Fig. 2). Triangle $B H C$ is a right isosceles triangle, so $C H=B H$. Triangles $A B H$ and $B D H$ are right triangles with an angle of $30^{\circ}$, so in them the legs opposite the angle of $30^{\circ}$ are half the hypotenuse. It turns out that $2 B H=A B, 2 H D=B D$.

Then

$$
C D=C H+H D=B H+H D=\frac{A B}{2}+\frac{B D}{2}=\frac{A B+A D}{2}=8
$$

Problem 4/3. On side $A C$ of triangle $A B C$, a point $D$ is chosen. It is known that $\angle B A C=30^{\circ}, \angle D B C=75^{\circ}, \angle B C A=45^{\circ}$. Find $C D$, if it is known that $B A+A D=18$.

Answer: 9.

Solution. From the sum of the angles of triangle $A B C$, it follows that $\angle A B D=30^{\circ}$. Therefore, $A D=B D$.

![](https://cdn.mathpix.com/cropped/2024_05_06_e1