Question

Arithmetic sequences $(a_n)$ and $(b_n)$ have integer terms with $a_1=b_1=1\lt a_2\le b_2$ and $a_nb_n=2010$ for some $n.$ What is the largest possible value of $n?$

$2$

$3$

$8$

$288$

$2009$

Answer: C

Concepts:arithmetic sequence divisibility greatest common divisor

Difficulty rating: 1950

Solution:

Since $a_n=1+(n-1)d_1$ and $b_n=1+(n-1)d_2$ for integers $d_1,d_2,$ the value $n-1$ divides both $a_n-1$ and $b_n-1,$ hence divides $\gcd(a_n-1,b_n-1).$

The factor pairs of $2010$ with $2\le a_n\le b_n$ are $(2,1005),$ $(3,670),$ $(5,402),$ $(6,335),$ $(10,201),$ $(15,134),$ and $(30,67).$

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

The sequences $a_n=2n-1$ and $b_n=19n-18$ realize this, so the largest value is $8.$

Thus, C is the correct answer.

2010 AMC 12A Problem 20

Answer: C

Difficulty rating: 1950

Solution:

Problem 20 in Other Years