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

2010 AMC 12A Problem 20

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20.

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

Small hint:

Write

a_n=1+(n-1)d_1

and

b_n=1+(n-1)d_2,

n-1

divides both

a_n-1

and

b_n-1

Big hint:

Check each factor pair of

2010

with

a_n\le b_n;

then

n-1

must divide

\gcd(a_n-1,b_n-1)

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,

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.