# 7. Clone 1

On the table, there are 14 coins of 2 and 5 rubles (each denomination is present). Some of them are face up with their denomination (tails) side, and some are face up with the heads side. If every coin on the table is flipped, the sum of the visible denominations will become three times greater than it was initially. How many 5-ruble coins can be on the table? List all possible options.

## Answers: 4 or 8
Solution. If we denote the sum of the visible denominations as $x$, then the sum of the denominations of the flipped coins will be $3x$, meaning the total amount of money on the table is $4x$, which is divisible by 4. Since both denominations are present, the total amount on the table can range from 31 to 67 rubles. The minimum amount is obtained by taking 1 coin of 5 rubles and 13 coins of 2 rubles: $5 + 2 \cdot 13 = 31$, and the maximum amount is obtained by taking 13 coins of 5 rubles and 1 coin of 2 rubles: $2 + 13 \cdot 5 = 67$. By replacing one 2-ruble coin with a 5-ruble coin, the total amount increases by 3 rubles. Therefore, the possible amounts are: $31, 34, 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67$.

Among these numbers, only 40, 52, and 64 are divisible by 4. We will show that 40 rubles could have been on the table. Suppose there are 10 coins of 2 rubles and 4 coins of 5 rubles. Initially, 5 two-ruble coins are face up, totaling 10 rubles, and after flipping, the total is 30 rubles.

We will also show that 52 rubles could have been on the table. Suppose there are 6 coins of 2 rubles and 8 coins of 5 rubles. Initially, 4 two-ruble coins and 1 five-ruble coin are face up, totaling 13 rubles. After flipping, the total is 39 rubles.

An example for 64 rubles does not exist, as the coins must be divided into groups with sums of 16 and 48, which is impossible with 14 coins. We will demonstrate this. Consider how to get 16 rubles. We must take an even number of 5-ruble coins, otherwise, we would need to make an odd sum with 2-ruble coins, which is impossible. Therefore, only two cases are possible:

1) Take 2 coins of 5 rubles and 3 coins of 2 rubles, then we have 9 coins left, which give a sum of 45 rubles $< 48$ rubles.
2) Take 8 coins of 2 rubles, then we have 6 coins left, with which we can get a sum of no more than 30 rubles $< 48$ rubles.

Note. In the 3 and 4 clones of this problem, the solution idea is the same. But all three examples can be constructed.

## Clone 2

On the table, there are 14 coins of 2 and 5 rubles (each denomination is present). Some of them are face up (tails up), and some are face down (heads up). If each coin on the table is flipped, the sum of the visible denominations will become 3 times greater than it was initially. How many 2-ruble coins can be on the table? List all possible options.

## Answers: 6 or 10

## Clone 3

On the table, there are 13 coins of 2 and 5 rubles (each denomination is present). Some of them are face up (tails up), and some are face down (heads up). If each coin on the table is flipped, the sum of the visible denominations will become 3 times greater than it was initially. How many 2-ruble coins can be on the table? List all possible options.

## Clone 4

On the table, there are 13 coins of 2 and 5 rubles (each denomination is present). Some of them are face up (tails up), and some are face down (heads up). If each coin on the table is flipped, the sum of the visible denominations will become 3 times greater than it was initially. How many 5-ruble coins can be on the table? List all possible options.

## Answers: 10 or 6 or 2