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2002 AMC 12A Problem 21

Problem 21 of 25HarderAlgebraNumber TheoryProblem-Solving Techniques

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

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Solution

Continuing the sequence gives 4,7,1,8,9,7,6,4, 7, 1, 8, 9, 7, 6, 3,9,2,1,3,4,7,1,…,3, 9, 2, 1, 3, 4, 7, 1, \ldots, which repeats with period 12.12. Each block of 1212 terms sums to 60.60. The largest kk with 60k≤10,00060k \le 10{,}000 is k=166,k = 166, giving S12⋅166=9960.S_{12\cdot 166} = 9960. Adding the next terms 4,7,1,8,9,7,64, 7, 1, 8, 9, 7, 6 contributes 42,42, pushing the total past 10,000.10{,}000. So n=12⋅166+7=1999.n = 12\cdot 166 + 7 = 1999. Thus, the correct answer is B.
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Tagged: units digit · summation · pattern recognition

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