2018 AMC 10B Problem 8

The number of toothpicks does not grow at a steady rate, so count them by their pattern.

1. Split into horizontal and vertical. The staircase looks the same if you flip it over its diagonal, so the horizontal and vertical toothpicks are equal in number. In the picture, $18$ toothpicks means $9$ horizontal ones.
2. Count horizontal toothpicks column by column. In the $3$-step staircase the columns are $3$, $2$ and $1$ squares tall. A column of $k$ squares needs $k + 1$ horizontal toothpicks (one under each square plus one on top). So the columns need $4 + 3 + 2 = 9$. That matches.
3. An $n$-step staircase has columns $n, n-1, \ldots, 1$ squares tall, so its horizontal toothpicks number
$$(n+1) + n + \cdots + 3 + 2.$$
4. Use the total. $180$ toothpicks means $90$ horizontal ones, so
$$2 + 3 + \cdots + (n+1) = 90.$$
Add $1$ to both sides to get a full sum starting from $1$:
$$1 + 2 + 3 + \cdots + (n+1) = 91.$$
5. Find where the sum reaches 91. The sum $1 + 2 + \cdots + m$ equals $\frac{m(m+1)}{2}$. Since $\frac{13 \times 14}{2} = 91$, we need $n + 1 = 13$, so $n = 12$.

Check: a $12$-step staircase has $2 \times 90 = 180$ toothpicks.

The answer is $\boxed{\textbf{(C)}\ 12}$.

Both the number of columns and the height of the tallest column grow with the number of steps, so the total grows faster and faster. Finding a formula from the structure, and checking it against the given $3$-step picture, protects you from guessing a steady rate.

Build a table of small cases: $1$ step uses $4$ toothpicks, $2$ steps use $10$, $3$ steps use $18$. These are $1 \times 4$, $2 \times 5$, $3 \times 6$, so $n$ steps use $n \times (n+3)$. Now $12 \times 15 = 180$, so $n = 12$.

Count the horizontal toothpicks of an $n$-step staircase column by column. The tallest column is a stack of $n$ squares and needs $n + 1$ horizontal toothpicks; the next column needs $n$, and so on down to the last column, which needs $2$. So the horizontals number
$$(n+1) + n + \cdots + 2 = \frac{(n+1)(n+2)}{2} - 1.$$
The staircase is symmetric about its diagonal, so the vertical toothpicks give the same count. Total:
$$2\left[\frac{(n+1)(n+2)}{2} - 1\right] = (n+1)(n+2) - 2 = n^2 + 3n = n(n+3).$$
Check: $n = 3$ gives $3 \cdot 6 = 18$, matching the figure.

Now solve $n(n+3) = 180$. Since $12 \cdot 15 = 180$, $n = 12$ (the other root, $-15$, is negative).

The answer is $\boxed{\textbf{(C)}\ 12}$.