2015 AMC 10A Problem 3

Count the horizontal toothpicks in the $3$-step staircase. The bottom row of squares has $3$ toothpicks on its floor and $3$ on its ceiling; the next row up adds a ceiling of $2$; the top row adds a ceiling of $1$. That is $3 + 3 + 2 + 1 = 9$ horizontal toothpicks. By the symmetry of the figure there are also $9$ vertical ones, for $18$ total, matching the problem.

For a $5$-step staircase the same count gives $5 + 5 + 4 + 3 + 2 + 1 = 20$ horizontal toothpicks and $20$ vertical, so $40$ in all.

Ann already has $18$, so she must add $40 - 18 = 22$.

The answer is $\boxed{\textbf{(D)}\ 22}$.

A staircase of $n$ steps has $n + (n + (n-1) + \cdots + 1) = n + \frac{n(n+1)}{2}$ toothpicks in each direction, so $n(n+3)$ total: $18$ for $n = 3$ and $40$ for $n = 5$. Verifying the formula against the given $18$ is a free sanity check before applying it to the new case.

Count what each new step adds. Going from $n$ steps to $n + 1$ steps adds a new diagonal of $n + 1$ squares, each needing $2$ new toothpicks, plus $2$ more for the new bottom-left corner: $2(n+1) + 2 = 2n + 4$ toothpicks. From $3$ to $4$ steps adds $10$, from $4$ to $5$ adds $12$. Total added: $22$.

• Counting the toothpicks in the whole $5$-step staircase ($40$) and forgetting to subtract the $18$ already placed.
• Assuming the toothpick count grows linearly (adding $6$ per step gives $30$ for five steps and the wrong answer $12$).