# 8. Variant 1.

Identical coins are laid out on a table in the shape of a hexagon. If they are laid out so that the side of the hexagon consists of 2 coins, then 7 coins are enough, and if the side consists of 3 coins, then a total of 19 coins are required. How many coins are needed to build a hexagon with a side consisting of 10 coins?
![](https://cdn.mathpix.com/cropped/2024_05_06_f37fbb269e2072e98fe4g-6.jpg?height=248&width=470&top_left_y=724&top_left_x=774)
Answer: 271.

Solution.

Method 1.

We will divide the coins into layers (contours) from the center. The first layer contains 1 coin, the second layer contains 6, and so on. Notice that each new layer contains 6 more coins than the previous one (if we remove the coins at the vertices, we get exactly as many coins as there were in the previous layer). Therefore, the total number of coins can be calculated using the formula $1+6+12+18+24+30+36+42+48+54=271$.

Method 2.

Let the side of the hexagon contain $n$ coins. Consider two opposite sides. Each of them consists of $n$ coins. To each of these sides, there are 2 more sides attached. But we have already counted one coin from each of these sides, so there are $n-1$ coins left on each of these sides, and two coins are counted twice. In total, there will be $2n + 4(n-1) - 2 = 6n - 6$. It remains to notice that each new layer of coins is built around the existing one. Then the total will be: $19 + (6 \cdot 4 - 6) + (6 \cdot 5 - 6) + (6 \cdot 6 - 6) + (6 \cdot 7 - 6) + (6 \cdot 8 - 6) + (6 \cdot 9 - 6) + (6 \cdot 10 - 6) = 19 + 18 + 24 + 30 + 36 + 42 + 48 + 54 = 271$.