2002 AMC 12A Problem 21

Question

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

Write out terms until the pattern repeats

The first

12

terms sum to

60

and then repeat; find how many full blocks stay under

10{,}000

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.

Problem 21 in other years