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

Latest revision as of 16:41, 26 August 2024

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; 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.)

Solution 3 (using answer choices)

Consider $n=288$ , which would imply $b_{288}\ge a_{288}\ge 288$ . However then $a_n b_n\ge 288^2>2010$ , so we just need to show that $n=8$ is achievable. This is true when $a_n=1+2n$ and $b_n=1+19n$ , giving $a_8 b_8=(15)(134)=2010$ . Hence the answer is $\boxed{\textbf{(C)}\ 8}$ .

Alternative Thinking

Since

$a_n*b_n = 2010,$

and

$a_n \le b_n$ , blue+yellow=green

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 $a_n = 2, 3, 5, 6, 112, 15,$ 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 $8 \Rightarrow \boxed{\text{C}}$ .

@@ Line 25: / Line 25: @@
-Hence, we have to find the largest <math>n</math> such that <math>\frac{a_n-1}{n-1}</math> and <math>\frac{b_n-1}{n-1}</math> are both integers.
+Hence, we have to find the largest <math>n</math> such that <math>\frac{a_n-1}{n-1}</math> and <math>\frac{b_n-1}{n-1}</math> are both integers; equivalently, we want to maximize <math>\gcd(a_n-1, b_n-1)</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>
+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>, noting that these are the possible values of <math>(a_n, b_n)</math> and we need <math>a_n \leq b_n</math>:
 <cmath>(2,1005), (3, 670), (5,402), (6,335), (10,201),(15,134),(30,67)</cmath>
@@ Line 46: / Line 46: @@
 Hence <math>n=8</math> is the largest possible <math>n</math>. (There is no need to check <math>n=2</math> anymore.)
-==Solution 3==
+=== Solution 3 (using answer choices) ===
-Using answer choices we see that the last two are way too big so we check 8 first and we finish it in like ~1 minute unlike the [censored] above who actually thought about the problem.
+Consider <math>n=288</math>, which would imply <math>b_{288}\ge a_{288}\ge 288</math>. However then <math>a_n b_n\ge 288^2>2010</math>, so we just need to show that <math>n=8</math> is achievable. This is true when <math>a_n=1+2n</math> and <math>b_n=1+19n</math>, giving <math>a_8 b_8=(15)(134)=2010</math>. Hence the answer is <math>\boxed{\textbf{(C)}\ 8}</math>.
+==Alternative Thinking==
+Since
+<math>a_n*b_n = 2010,</math>
+and
+<math>a_n \le b_n</math>,
+blue+yellow=green
+it follows that
+<math>a_n \le \sqrt{2010} \Rightarrow a_n \le 44</math>.
+But <math>a_n</math> and <math>b_n</math> are also integers, so <math>a_n</math> must be a factor of <math>2010</math> smaller than <math>44</math>. Notice that <math>2010 = 2*3*5*67</math>. Therefore <math>a_n = 2, 3, 5, 6, 112, 15,</math> or <math>30</math> and <math>b_n = 1005, 670, 402, 335, 201, 134,</math> or <math>67</math>; respectively.
+Notice that the term <math>a_m</math> is equivalent to the first term <math>a_1 = 1</math> plus <math>(m-1)</math> times the common difference for that particular arithmetic sequence. Let the common difference of <math>(a_n)</math> be <math>k</math> and the common difference of <math>(b_n)</math> be <math>i</math> (not <math>\sqrt{-1}</math>). Then
+<math>a_n</math> (the <math>n</math>th term, not the sequence itself) <math>=1 + k(n-1)</math>
+and
+<math>b_n = 1 + i(n-1)</math>
+Subtracting one from all the possible values listed above for <math>a_n</math> and <math>b_n</math>, we get
+<math>k(n-1) = 1, 2, 4, 5, 9, 14, 29</math>
+and
+<math>i(n-1) = 1004, 669, 401, 334, 200, 133, 66</math>
+In order to maximize <math>n</math>, we must maximize <math>n-1</math>. Therefore <math>k</math> and <math>i</math> are [[coprime]] and <math>n-1</math> is the [[Greatest common factor|GCF]] of any corresponding pair. Inspecting all of the pairs, we see that the [[Greatest common factor|GCF]] is always <math>1</math> except for the pair <math>(14, 133),</math> which has a GCF of <math>7</math>. Therefore the maximum value of <math>n</math> is <math>8 \Rightarrow \boxed{\text{C}}</math>.
 == 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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