Rynbniov i...

In the store, 20 kg of cheese was delivered, and a queue formed. After selling cheese to the next customer, the saleswoman accurately calculates the average weight of the purchase for all the cheese sold and informs how many people the remaining cheese will last if everyone buys exactly this average weight. Could the saleswoman have reported after each of the first 10 customers that the cheese would last exactly 10 more people? If so, how much cheese was left in the store after the first 10 customers?

#
Let $s_{k}$ be the average weight of cheese sold to the first $k$ customers. By the condition $20-k s_{k}=10 s_{k}$, hence

$s_{k}=\frac{20}{k+10}$ kg, and after the $k$-th customer, there remains

200

$k+10$

kg of cheese. Since the obtained sequence is decreasing, the seller can make such a statement after each customer (not limited to the first ten). After the 10th customer, exactly half of the cheese remains - 10 kg.

## [ Problem 98538 Topics: [ Continuity considerations ]  [ Examples and counterexamples. Constructions ]  Difficulty: $3+$ Grades: $9,10,11$

Author: G. Talperin $[$. .

There exist 1000 consecutive natural numbers, none of which are prime (for example, $1001!+2, 1001!+3, \ldots, 1001!+1001$ ).

But do there exist 1000 consecutive natural numbers, among which there are exactly five prime numbers?

## Solution

Let $P_{n}$ denote the number of prime numbers among a thousand consecutive numbers from $n$ to $n+999$. Note that $P_{n+1}$ differs from $P_{n}$ by no more than one. However, $P_{1}>5$, and $P_{1001!+2}=0<5$, so when $n$ changes from 1 to $1001!+2$, the value of $P_{n}$ will be equal to 5 for some $n$.

## Answer

They exist.