2010 AMC 12A Problem 20

Below is the professionally curated solution for Problem 20 of the 2010 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 12A solutions, or check the answer key.

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Concepts:arithmetic sequencedivisibilitygreatest common divisor

Difficulty rating: 1950

20.

Arithmetic sequences (an)(a_n) and (bn)(b_n) have integer terms with a1=b1=1<a2b2a_1=b_1=1\lt a_2\le b_2 and anbn=2010a_nb_n=2010 for some n.n. What is the largest possible value of n?n?

22

33

88

288288

20092009

Solution:

Since an=1+(n1)d1a_n=1+(n-1)d_1 and bn=1+(n1)d2b_n=1+(n-1)d_2 for integers d1,d2,d_1,d_2, the value n1n-1 divides both an1a_n-1 and bn1,b_n-1, hence divides gcd(an1,bn1).\gcd(a_n-1,b_n-1).

The factor pairs of 20102010 with 2anbn2\le a_n\le b_n are (2,1005),(3,670),(5,402),(6,335),(10,201),(15,134),(2,1005),(3,670),(5,402),(6,335), (10,201),(15,134), and (30,67).(30,67).

For every pair except (15,134),(15,134), the numbers an1a_n-1 and bn1b_n-1 are relatively prime, forcing n=2.n=2. For (15,134),(15,134), gcd(14,133)=7,\gcd(14,133)=7, so n1n-1 can equal 7,7, giving n=8.n=8.

The sequences an=2n1a_n=2n-1 and bn=19n18b_n=19n-18 realize this, so the largest value is 8.8.

Thus, C is the correct answer.

Problem 20 in Other Years