2010 AMC 12A Problems/Problem 20

Problem 20

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

$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 288 \qquad \textbf{(E)}\ 2009$

Solution

Since $\left(a_n\right)$ and $\left(b_n\right)$ have integer terms with $a_1=b_1=1$ , we can write the terms of each sequence as

$\left(a_n\right) \Rightarrow \{1, x+1, 2x+1, 3x+1, ...\}$

$\left(b_n\right) \Rightarrow \{1, y+1, 2y+1, 3y+1, ...\}$

where $x$ and $y$ are the common differences of each, respectively.

Since

$a_n = (n-1)x+1$

$b_n = (n-1)y+1$

it is easy to see that

$a_n \equiv b_n \equiv 1 \mod{(n-1)}$ .

Hence, we have to find the largest $n$ such that $\frac{a_n-1}{n-1}$ and $\frac{b_n-1}{n-1}$ are both integers.

The prime factorization of $2010$ is $2\cdot{3}\cdot{5}\cdot{67}$ . We list out all the possible pairs that have a product of $2010$

$(2,1005), (3, 670), (5,402), (6,335), (10,201),(15,134),(30,67)$

and soon find that the largest $n-1$ value is $7$ for the pair $(15, 134)$ , and so the largest $n$ value is $\boxed{8\ \textbf{(C)}}$ .

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2010 AMC 12A Problems/Problem 20

Problem 20

Solution