2018 AMC 10B Problem 8

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 $\boxed{\textbf{(C)}\ 12}$.

Toothpick and tile patterns are almost always quadratic in the step number because both the number of pieces per row and the number of rows grow. Derive the formula from the structure (or from small cases), verify it against the given example, then solve; the given "3 steps, 18 toothpicks" is a built-in check.

Small cases: $1$ step uses $4$, $2$ steps use $10$, $3$ steps use $18$. These are $1 \cdot 4$, $2 \cdot 5$, $3 \cdot 6$, suggesting $n(n+3)$, and $180 = 12 \cdot 15$ fits the pattern. Alternatively, differences $6, 8, 10, \dots$ increase by $2$, confirming a quadratic.

• Assuming the count is proportional to the number of steps (18/3 = 6 per step, so 30 steps), when it actually grows quadratically.
• Counting $n$ horizontal toothpicks per column instead of $n + 1$ (forgetting the bottom or top edge), which gives $n(n+1)$ and no integer solution to $n(n+1) = 180$.