1994 AMC 12 Problem 6

Attempt Problem 6 of the 1994 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1994 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

6.

In the sequence …,a,b,c,d,0,1,1,2,3,5,8,… \ldots,a,b,c,d,0,1,1,2,3,5,8,\ldots each term is the sum of the two terms to its left. Find a.a.
−3-3
−1-1
00
11
33
Answer: A
Concepts:Fibonacci recurrenceworking backward
Small hint:
Work backward by subtracting the preceding known term
Big hint:
Determine d,c,bd,c,b in that order before finding aa
Solution:
Working backward, d+0=1d+0=1 gives d=1,d=1, then c+d=0c+d=0 gives c=−1,c=-1, and b+c=db+c=d gives b=2.b=2. Finally a+b=c,a+b=c, so a=−1−2=−3.a=-1-2=-3. Thus the correct answer is A.
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