2015 AMC 10A Problem 3

First learn how to count the $3$-step staircase and check the count against the $18$ given. Then use the same count for $5$ steps.

1. Flat (horizontal) toothpicks, 3 steps. The bottom row of squares is $3$ squares wide. It needs $3$ toothpicks along its floor and $3$ along its top. The row above is $2$ squares wide and adds a top of $2$. The top row adds a top of $1$. That is
$$3 + 3 + 2 + 1 = 9.$$
2. Standing (vertical) toothpicks. The staircase looks the same if you flip it along its diagonal, so there are also $9$ standing toothpicks.
3. Total: $9 + 9 = 18$. This matches the problem, so the counting method is right.
4. Now 5 steps. Flat toothpicks: $5$ along the floor, then tops of $5, 4, 3, 2, 1$:
$$5 + 5 + 4 + 3 + 2 + 1 = 20.$$
Standing toothpicks: also $20$. Total: $40$.
5. Ann already has $18$ toothpicks in place, so she needs $40 - 18 = 22$ more.

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

Counting a grid figure is easier when you split the pieces into two directions and count each direction row by row. Checking your method against the given picture ($18$) before using it on the new case catches mistakes for free. In general an $n$-step staircase uses $n(n + 3)$ toothpicks: $18$ when $n = 3$ and $40$ when $n = 5$.

Count what each new step adds. Going from $3$ steps to $4$ steps adds a new diagonal of $4$ squares. Each new square needs $2$ new toothpicks, and the new bottom corner needs $2$ more, so that adds $4 \times 2 + 2 = 10$. Going from $4$ steps to $5$ steps adds $5 \times 2 + 2 = 12$ in the same way. In total Ann adds $10 + 12 = 22$.