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

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== Problem 20 ==
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== Problem ==
 
Arithmetic sequences <math>\left(a_n\right)</math> and <math>\left(b_n\right)</math> have integer terms with <math>a_1=b_1=1<a_2 \le b_2</math> and <math>a_n b_n = 2010</math> for some <math>n</math>. What is the largest possible value of <math>n</math>?
 
Arithmetic sequences <math>\left(a_n\right)</math> and <math>\left(b_n\right)</math> have integer terms with <math>a_1=b_1=1<a_2 \le b_2</math> and <math>a_n b_n = 2010</math> for some <math>n</math>. What is the largest possible value of <math>n</math>?
  
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Since <math>\left(a_n\right)</math> and <math>\left(b_n\right)</math> have integer terms with <math>a_1=b_1=1</math>, we can write the terms of each sequence as
 
Since <math>\left(a_n\right)</math> and <math>\left(b_n\right)</math> have integer terms with <math>a_1=b_1=1</math>, we can write the terms of each sequence as
  
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<cmath>\begin{align*}&\left(a_n\right) \Rightarrow \{1, x+1, 2x+1, 3x+1, ...\}\\
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<math>\left(a_n\right) \Rightarrow \{1, x+1, 2x+1, 3x+1, ...\}</math>
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&\left(b_n\right) \Rightarrow \{1, y+1, 2y+1, 3y+1, ...\}\end{align*}</cmath>
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<math>\left(b_n\right) \Rightarrow \{1, y+1, 2y+1, 3y+1, ...\}</math>
 
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where <math>x</math> and <math>y</math> (<math>x\leq y</math>) are the common differences of each, respectively.
 
where <math>x</math> and <math>y</math> (<math>x\leq y</math>) are the common differences of each, respectively.
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Since
 
Since
  
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<math>a_n = (n-1)x+1</math>
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<cmath>\begin{align*}a_n &= (n-1)x+1\\
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b_n &= (n-1)y+1\end{align*}</cmath>
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<math>b_n = (n-1)y+1</math>
 
  
 
it is easy to see that
 
it is easy to see that
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The prime factorization of <math>2010</math> is <math>2\cdot{3}\cdot{5}\cdot{67}</math>. We list out all the possible pairs that have a product of <math>2010</math>
 
The prime factorization of <math>2010</math> is <math>2\cdot{3}\cdot{5}\cdot{67}</math>. We list out all the possible pairs that have a product of <math>2010</math>
  
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<cmath>(2,1005), (3, 670), (5,402), (6,335), (10,201),(15,134),(30,67)</cmath>
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<math>(2,1005), (3, 670), (5,402), (6,335), (10,201),(15,134),(30,67)</math>
 
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and soon find that the largest <math>n-1</math> value is <math>7</math> for the pair <math>(15, 134)</math>, and so the largest <math>n</math> value is <math>\boxed{8\ \textbf{(C)}}</math>.
 
and soon find that the largest <math>n-1</math> value is <math>7</math> for the pair <math>(15, 134)</math>, and so the largest <math>n</math> value is <math>\boxed{8\ \textbf{(C)}}</math>.
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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.)
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== See also ==
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{{AMC12 box|year=2010|num-b=19|num-a=21|ab=A}}
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[[Category:Introductory Algebra Problems]]

Revision as of 22:32, 25 February 2010

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.)

See also

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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