Difference between revisions of "2010 AMC 12A Problems/Problem 20"

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Hence <math>n=8</math> is the largest possible <math>n</math>. (There is no need to check <math>n=2</math> anymore.)
Hence <math>n=8</math> is the largest possible <math>n</math>. (There is no need to check <math>n=2</math> anymore.)
==Alternative Thinking==
==Alternative Thinking==

Latest revision as of 21:25, 23 November 2020


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 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.


\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; equivalently, we want to maximize $\gcd(a_n-1, b_n-1)$.

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 $2010$, noting that these are the possible values of $(a_n, b_n)$ and we need $a_n \leq b_n$:

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

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


$a_n*b_n = 2010,$


$a_n \le b_n$,

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) $=1 + k(n-1)$


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

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

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


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

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 $8 \Rightarrow \boxed{\text{C}}$.

See also

2010 AMC 12A (ProblemsAnswer KeyResources)
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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