2015 AMC 10A Problem 7

Count the jumps first, then turn jumps into terms.

1. Each term is $3$ more than the one before: $13, 16, 19, \ldots$
2. From the first term $13$ to the last term $73$ is a climb of $73 - 13 = 60$.
3. Each jump is $3$, so there are $60 \div 3 = 20$ jumps.
4. Think of a tiny case: the list $13, 16, 19$ has $2$ jumps and $3$ numbers. There is always one more number than jumps, because the first number comes before any jump.
5. So $20$ jumps give $20 + 1 = 21$ terms.

The answer is $\boxed{\textbf{(B)}\ 21}$.

This is the fence-post idea: $20$ gaps need $21$ posts. For any list that goes up by the same step, the number of terms is $\frac{\text{last} - \text{first}}{\text{step}} + 1$. Trying a tiny case, like three numbers, is a quick way to remember whether to add $1$.

Change the list into one you can count at a glance. Subtract $10$ from every term to get $3, 6, 9, \ldots, 63$. Then divide every term by $3$ to get $1, 2, 3, \ldots, 21$. Neither step changes how many numbers there are, so there are $21$.

The common difference is $3$. The $n$-th term is $13 + 3(n - 1)$, and it equals $73$ when
$$3(n - 1) = 60 \quad\Longrightarrow\quad n - 1 = 20 \quad\Longrightarrow\quad n = 21.$$

Equivalently, there are $20$ jumps of size $3$ from $13$ to $73$, and $20$ jumps connect $21$ terms.

The answer is $\boxed{\textbf{(B)}\ 21}$.