How many coins are needed to make as many rows of one coin as there are coins? The answer is 1: you make

**1** row of one coin using that

**1** coin.

How many coins are needed for rows of two coins? The answer is 3: put the coins in a triangle and you make **3** rows of two coins using **3** coins.

PART 1

Now, how many coins are needed to make as many rows of 3 as there are coins?

PART 2

If that was easy, how many for 4?

Note: if you are making rows of 4, any line drawn can intersect with a maximum of 4 coins. (you cannot place 5 coins in row and count it as two rows of 4) And no stacking coins.

(In reply to

First Part by TamTam)

TamTam:

I count two more diagonals in your diagram, for a total of 13, but you
can certainly rearrange it so that those two diagonals are not linear,
giving the 11.

There is an question of whether o o o o is zero or two or four rows of
three. You apparently counted contiguous linear coins, but not
non-contiguous linear coins, resulting in two. This is
reasonable, but not the only way to count.

If we count this way, though, then I come up with 8 as a minimum. My Layout:

1 2 3

4

5

6 7

8

Where 4 is at the intersection of the lines joining 2 to 7 and 3 to 5.

The 8 rows are 123, 157, 168, 345, 456, 378, 258, and 247.