Shapovalov A.V.

In a sequence of natural numbers, each number, except the first, is obtained by adding to the previous one its largest digit.

What is the maximum number of consecutive odd terms that can occur in the sequence?
Let $a_{1}, \ldots, a_{5}$ be consecutive odd terms of a sequence. The last digits of these numbers are odd, and the largest digits of the numbers $a_{1}, \ldots, a_{4}$ are even and, therefore, not the last. When moving to the next number, the largest digit of the previous number does not change (otherwise, it would increase by 1 and become odd), so $a_{1}, \ldots, a_{5}$ form an arithmetic progression with a non-zero even digit $d$ as the common difference. The numbers $0, d, 2d, 3d, 4d$ end in different digits (since they give different remainders when divided by 5). By adding these numbers to $a_{1}$, we see that $a_{1}, \ldots, a_{5}$ also end in different digits. Therefore, one of the numbers must end in

9. But this can only be $a_{5}$, and thus, the next term in the sequence is an even number.

Five consecutive odd terms can be found, for example, in the sequence: 807, 815, 823, 831, 839.

## Answer

5 terms.

The area of triangle $ABC$ is $10 \text{ cm}^2$. What is the smallest value that the radius of the circumscribed circle of triangle $ABC$ can take, given that the midpoints of the heights of this triangle lie on the same line?

## Solution

According to the solution of problem 102505, triangle $ABC$ is a right triangle. Let its legs be $a$ and $b$, the hypotenuse be $c$, and the radius of the circumscribed circle be $R$. We can reason in different ways.

First method. $4R^2 = c^2 = a^2 + b^2 \geq 2ab = 40$. Therefore, $R^2 \geq 10$.

Second method. From the solution of problem 102505, we see that $10 = S_{ABC} \leq R^2$.

Equality is achieved for an isosceles right triangle.

## Answer

$\sqrt{10}$.