## Task B-1.7.

The arithmetic mean of a sequence of numbers $x_{1}, x_{2}, \ldots, x_{n}$ is 100. The largest of these numbers is a three-digit number $x_{n}$. If it is replaced with $4 x_{n}$, the arithmetic mean of the new sequence of numbers is 200. What is the maximum number of numbers that can be in such a sequence? What is the value of the number $x_{n}$ in this case?
## Solution.

From the conditions of the problem, we have:

$$
\begin{aligned}
\frac{x_{1}+x_{2}+\cdots+x_{n}}{n} & =100 \\
\frac{x_{1}+x_{2}+\cdots+4 x_{n}}{n} & =200
\end{aligned}
$$

Then,

$$
\begin{aligned}
x_{1}+x_{2}+\cdots+x_{n} & =100 n \\
x_{1}+x_{2}+\cdots+4 x_{n} & =200 n
\end{aligned}
$$

If we subtract the first equation from the second, we get \(3 x_{n}=100 n\).

Since \(x_{n}\) is a three-digit natural number, it follows that the number \(n\) is divisible by 3

and the number \(\frac{100 n}{3}\) is a three-digit number.

This is true for \(n \in\{3,6,9,12,15,18,21,24,27\}\). 1 point

Since we are looking for the largest number \(n\) with the given property, then \(n=27\)

and we have \(x_{n}=\frac{100 n}{3}=\frac{100 \cdot 27}{3}=900\).

## SCHOOL/CITY COMPETITION IN MATHEMATICS

## 2nd grade - high school - B variant

28. January 2019.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.