2002 AMC 12A Problem 21

Problem 21 of 25HarderAlgebra Number Theory

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

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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

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.

n = 12\cdot 166 + 7 = 1999.

Thus, the correct answer is B.