2002 AMC 12A Problem 21

Below is the professionally curated solution for Problem 21 of the 2002 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 12A solutions, or check the answer key.

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Concepts:units digit summation pattern recognition

Difficulty rating: 1840

21.

Consider the sequence of numbers $4, 7, 1, 8, 9, 7, 6, \ldots$ For $n \gt 2,$ the $n$ th term of the sequence is the units digit of the sum of the two previous terms. Let $S_n$ denote the sum of the first $n$ terms of this sequence. The smallest value of $n$ for which $S_n \gt 10{,}000$ is

$1992$

$1999$

$2001$

$2002$

$2004$

Solution:

Continuing the sequence gives $4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3, 4, 7, 1, \ldots,$ which repeats with period $12.$ Each block of $12$ terms sums to $60.$

The largest $k$ with $60k \le 10{,}000$ is $k = 166,$ giving $S_{12\cdot 166} = 9960.$ Adding the next terms $4, 7, 1, 8, 9, 7, 6$ contributes $42,$ pushing the total past $10{,}000.$ So $n = 12\cdot 166 + 7 = 1999.$

Thus, the correct answer is B.

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