2003 AMC 10A Problem 23

Toothpicks are shared between neighboring triangles, so counting "3 per triangle" would count many twice. Find a way to give each toothpick to exactly one triangle.

1. Count the rows. In the picture the rows have $1$, $3$, $5$ small triangles: row $k$ has $2k - 1$. A bottom row of $2003$ triangles means $2k - 1 = 2003$, so there are $k = 1002$ rows.
2. Up and down triangles. In each row the triangles alternate pointing up and pointing down, starting and ending with up. Row $k$ has $k$ up-triangles and $k - 1$ down-triangles. (Row $3$ in the picture: $3$ up and $2$ down, which makes $5$.)
3. Each toothpick belongs to one up-triangle. Look at the picture. Every horizontal toothpick is the bottom edge of an up-triangle. Every slanted toothpick is the left or right side of an up-triangle. The down-triangles have no toothpicks of their own: their three sides are borrowed from the up-triangles around them. And no two up-triangles share a side.
4. So toothpicks $= 3 \times$ (number of up-triangles). Check on the picture: $1 + 2 + 3 = 6$ up-triangles, and $3 \times 6 = 18$ toothpicks, which matches counting them directly.
5. Count up-triangles for $1002$ rows.
$$1 + 2 + \cdots + 1002 = \frac{1002 \times 1003}{2} = 501 \times 1003 = 502503.$$
6. Toothpicks. $3 \times 502503 = 1507509$.

The answer is $\boxed{\textbf{(C)}\ 1,507,509}$.

When pieces are shared between shapes, look for a family of shapes that uses every piece exactly once. Here the up-pointing triangles do that, so the count becomes a single multiplication. Testing the rule on the small picture first catches the two usual mistakes: miscounting the rows and counting inside edges twice.

Count small cases and look for the pattern. With $1, 2, 3, 4$ rows you need $3, 9, 18, 30$ toothpicks, which is $3$ times $1, 3, 6, 10$ (the sums $1$, $1+2$, $1+2+3$, $1+2+3+4$). For $1002$ rows that is $3 \times 502503$. You do not even need to finish the multiplication: $3 \times 3$ ends in $9$, and only choice (C) ends in $9$.