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

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== Solution ==
 
== Solution ==
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=== Solution 1 ===
 
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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where <math>x</math> and <math>y</math> are the common differences of each, respectively.
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where <math>x</math> and <math>y</math> (<math>x\leq y</math>) are the common differences of each, respectively.
  
  
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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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=== Solution 2 ===
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As above, let <math>a_n=(n-1)x+1</math> and <math>b_n=(n-1)y+1</math> for some <math>1\leq x\leq y</math>.
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Now we get <math>2010 = a_n b_n = (n-1)^2xy + (n-1)(x+y) + 1</math>, hence <math>2009 = (n-1)( (n-1)xy + x + y )</math>. Therefore <math>n-1</math> divides <math>2009 = 7^2 \cdot 41</math>. And as the second term is greater than the first one, we only have to consider the options <math>n-1\in\{1,7,41\}</math>.
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For <math>n=42</math> we easily see that for <math>x=y=1</math> the right side is less than <math>49</math> and for any other <math>(x,y)</math> it is way too large.
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For <math>n=8</math> we are looking for <math>(x,y)</math> such that <math>7xy + x + y = 2009/7 = 7\cdot 41</math>. Note that <math>x+y</math> must be divisible by <math>7</math>. We can start looking for the solution by trying the possible values for <math>x+y</math>, and we easily discover that for <math>x+y=21</math> we get <math>xy + 3 = 41</math>, which has a suitable solution <math>(x,y)=(2,19)</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.)

Revision as of 07:46, 16 February 2010

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

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


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

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


where $x$ and $y$ ($x\leq y$) 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 $\boxed{8\ \textbf{(C)}}$.

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.)

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