2013 AMC 10B Problem 4

1. First see the rule on a tiny example. Count down from $10$ to $7$: you say $10, 9, 8, 7$. That is $4$ numbers, but $10 - 7 = 3$. You need one more than the difference, because both ends get said.
2. Test the rule on the fact given in the problem. Counting up from $3$, the number $53$ is said $53 - 3 + 1 = 51$st. That matches.
3. Now count down from $201$. The numbers said, up to and including $53$, are $201, 200, 199, \ldots, 54, 53$. How many is that?
$$201 - 53 + 1 = 149.$$
So $53$ is the $149$th number said.

The answer is $\boxed{\textbf{(D)}\ 149}$.

To count the whole numbers from $a$ to $b$, including both ends, use $b - a + 1$, not $b - a$. When unsure, check the rule on a list short enough to count by hand; the problem even gave you one such check.

Count the whole list first: from $3$ to $201$ there are $201 - 3 + 1 = 199$ numbers. Going forward, $50$ numbers come before $53$, so going backward, the numbers after $53$ are those $50$. The numbers before and including $53$ in the backward count are $199 - 50 = 149$.

Counting backwards from $201$, the numbers said before reaching $53$ are $201, 200, \dots, 54$, and then $53$ itself. The position of $53$ is the number of integers from $53$ to $201$ inclusive:
$$201 - 53 + 1 = 149.$$

The answer is $\boxed{\textbf{(D)}\ 149}$.