2010 AMC 12A Problems/Problem 20

Problem

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

Solution 1

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

$\begin{align*}&\left(a_n\right) \Rightarrow \{1, x+1, 2x+1, 3x+1, ...\}\\ &\left(b_n\right) \Rightarrow \{1, y+1, 2y+1, 3y+1, ...\}\end{align*}$

where $x$ and $y$ ( $x\leq y$ ) are the common differences of each, respectively.

Since

$\begin{align*}a_n &= (n-1)x+1\\ b_n &= (n-1)y+1\end{align*}$

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 $\boxed{8\ \textbf{(C)}}$ .

Solution 2

As above, let $a_n=(n-1)x+1$ and $b_n=(n-1)y+1$ for some $1\leq x\leq y$ .

Now we get $2010 = a_n b_n = (n-1)^2xy + (n-1)(x+y) + 1$ , hence $2009 = (n-1)( (n-1)xy + x + y )$ . Therefore $n-1$ divides $2009 = 7^2 \cdot 41$ . And as the second term is greater than the first one, we only have to consider the options $n-1\in\{1,7,41\}$ .

For $n=42$ we easily see that for $x=y=1$ the right side is less than $49$ and for any other $(x,y)$ it is way too large.

For $n=8$ we are looking for $(x,y)$ such that $7xy + x + y = 2009/7 = 7\cdot 41$ . Note that $x+y$ must be divisible by $7$ . We can start looking for the solution by trying the possible values for $x+y$ , and we easily discover that for $x+y=21$ we get $xy + 3 = 41$ , which has a suitable solution $(x,y)=(2,19)$ .

Hence $n=8$ is the largest possible $n$ . (There is no need to check $n=2$ anymore.)

Alternative Thinking

Since

$a_n*b_n = 2010,$

and

$a_n \le b_n$ ,

it follows that

$a_n \le \sqrt{2010} \Rightarrow a_n \le 44$ .

But $a_n$ and $b_n$ are also integers, so $a_n$ must be a factor of $2010$ smaller than $44$ . Notice that $2010 = 2*3*5*67$ . Therefore $a_n = 2, 3, 5, 6, 10, 15,$ or $30$ and $b_n = 1005, 670, 402, 335, 201, 134,$ or $67$ ; respectively.

Notice that the term $a_m$ is equivalent to the first term $a_1 = 1$ plus $(m-1)$ times the common difference for that particular arithmetic sequence. Let the common difference of $(a_n)$ be $k$ and the common difference of $(b_n)$ be $i$ (not $\sqrt{-1}$ ). Then

$a_n$ (the $n$ th term, not the sequence itself) $=1 + k(n-1)$

and

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

Subtracting one from all the possible values listed above for $a_n$ and $b_n$ , we get

$k(n-1) = 1, 2, 4, 5, 9, 14, 29$

and

$i(n-1) = 1004, 669, 401, 334, 200, 133, 66$

In order to maximize $n$ , we must maximize $n-1$ . Therefore $k$ and $i$ are coprime and $n-1$ is the GCF of any corresponding pair. Inspecting all of the pairs, we see that the GCF is always $1$ except for the pair $(14, 133),$ which has a GCF of $7$ . Therefore the maximum value of $n$ is $8 \Rightarrow \boxed{\text{C}}$ .

2010 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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