2000 AMC 10 Problem 6

The units digit of a sum is determined by the units digits of the addends, so we can run the recursion on last digits alone: each new digit is the previous two added, keeping only the ones place.

Starting $1, 1$:
$$1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, \ldots$$

Record the first appearance of each digit:

By the $15$th term, nine digits have shown up and only $6$ is missing. It finally appears as the $21$st term ($F_{21} = 10946$), so $6$ is the last digit to appear.

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

Reducing mod $10$ commutes with addition, so the sequence of last digits obeys the same recursion as the sequence itself and you never need the actual Fibonacci numbers, which quickly get large. A written checklist of which digits have appeared turns a vague "which comes last" question into a mechanical scan. Tagging note: the Fibonacci wording is surface; the work is arithmetic mod $10$.

Since the answer is one of the five listed digits, you only need to watch for $0, 4, 6, 7, 9$; the moment the last of those five is written, you are done. The list above shows $6$ trailing the other four.

• Stopping the list as soon as a tempting digit such as 0 or 7 appears, without confirming that every other digit has already occurred.
• An addition slip early in the list corrupts everything after it; cross-check the first terms against the known values $13, 21, 34, 55, 89, 144$.