2. Marko took all the coins out of his piggy bank and arranged them on the table in a single row. First, he placed the 2 euro coins, then he inserted one 50 cent coin between each two adjacent coins in the row. After that, he inserted a 20 cent coin between every two coins on the table. Finally, he inserted a 5 cent coin between every two coins in the new row. How many coins were arranged on the table, if their total value is 95 euros?
First solution.

First, 2-euro coins were placed on the table. Let there be $n$ of them.

So, there are $n$ coins worth $2 n$ euros on the table.

Between $n$ coins, $n-1$ new coins can be inserted.

Each is worth 50 cents or 0.5 euros, so the added amount is $(n-1) \cdot 0.5$ euros.

Now there are $n+n-1=2 n-1$ coins on the table,

with a value of $2 n+0.5 n-0.5=2.5 n-0.5$ euros.

Between the existing $2 n-1$ coins, another smaller coin can be placed, so $2 n-2$ coins.

These coins are worth 20 cents or 0.2 euros, so their value is $(2 n-2) \cdot 0.2=0.4 n-0.4$ euros.

Now there are $2 n-1+2 n-2=4 n-3$ coins on the table,

and their value is $2.5 n-0.5+0.4 n-0.4=2.9 n-0.9$ euros.

In the last round, between $4 n-3$ coins, $4 n-4$ coins can be placed.

Each is worth 5 cents or 0.05 euros, so the added amount is $(4 n-4) \cdot 0.05=0.2 n-0.2$ euros.

Finally, there are $4 n-3+4 n-4=8 n-7$ coins on the table,

with a total value of $2.9 n-0.9+0.2 n-0.2=3.1 n-1.1$ euros.

Since there are a total of 95 euros on the table, from $3.1 n-1.1=95$ we get

$3.1 n=95+1.1$

$3.1 n=96.1$

$n=96.1: 3.1$

$n=31$.

Now it is calculated that there are $8 \cdot 31-7=241$ coins on the table.