2000 AMC 10 Problem 6

1. The Fibonacci numbers get big fast, but we only care about the last digit. The last digit of a sum depends only on the last digits of the two numbers added. For example, $13 + 21 = 34$, and $3 + 1 = 4$ already tells us the last digit.
2. So work with last digits only. Each new digit is the sum of the previous two, keeping only the ones digit:
$$1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, \ldots$$
For example, after $5, 8$ comes $5 + 8 = 13$, so the digit $3$. After $8, 3$ comes $11$, so the digit $1$.
3. Keep a checklist of when each digit first shows up:

1. By term $15$, nine of the ten digits have appeared. Only $6$ is missing. It finally shows up at term $21$ (the actual number is $10946$).

So $6$ is the last digit to appear.

The answer is $\boxed{\textbf{(C)}\ 6}$.

The list of last digits follows the same "add the previous two" rule as the Fibonacci numbers, so you never need the large numbers themselves. Writing a checklist turns "which comes last" into a simple scan. Next time a problem asks only about last digits, drop everything except the ones digit.

The answer must be one of the five choices, so you only need to watch for $0, 4, 6, 7, 9$. In the list, $4$ appears at term $9$, $9$ at term $11$, $7$ at term $14$, $0$ at term $15$, and $6$ only at term $21$. So $6$ is last.