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Difference between revisions of "1985 AIME Problems/Problem 13"

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== Problem ==
 
== Problem ==
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The numbers in the sequence <math>101</math>, <math>104</math>, <math>109</math>, <math>116</math>,<math>\ldots</math> are of the form <math>a_n=100+n^2</math>, where <math>n=1,2,3,\ldots</math> For each <math>n</math>, let <math>d_n</math> be the greatest common divisor of <math>a_n</math> and <math>a_{n+1}</math>. Find the maximum value of <math>d_n</math> as <math>n</math> ranges through the positive integers.
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The numbers in the [[sequence]] <math>101</math>, <math>104</math>, <math>109</math>, <math>116</math>,<math>\ldots</math> are of the form <math>a_n=100+n^2</math>, where <math>n=1,2,3,\ldots</math> For each <math>n</math>, let <math>d_n</math> be the greatest common divisor of <math>a_n</math> and <math>a_{n+1}</math>. Find the maximum value of <math>d_n</math> as <math>n</math> ranges through the [[positive integer]]s.
  
 
== Solution ==
 
== Solution ==
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If <math>(x,y)</math> denotes the greatest common divisor of <math>x</math> and <math>y</math>, then we have <math>d_n=(a_n,a_{n+1})=(100+n^2,100+n^2+2n+1)</math>. Now assuming that <math>d_n</math> divides <math>100+n^2</math>, it must divide <math>2n+1</math> if it is going to divide the entire expression <math>100+n^2+2n+1</math>.
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If <math>(x,y)</math> denotes the [[greatest common divisor]] of <math>x</math> and <math>y</math>, then we have <math>d_n=(a_n,a_{n+1})=(100+n^2,100+n^2+2n+1)</math>. Now assuming that <math>d_n</math> [[divisor | divides]] <math>100+n^2</math>, it must divide <math>2n+1</math> if it is going to divide the entire [[expression]] <math>100+n^2+2n+1</math>.
  
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Thus the equation turns into <math>d_n=(100+n^2,2n+1)</math>. Now note that since <math>2n+1</math> is odd for integral <math>n</math>, we can multiply the left integer, <math>100+n^2</math>, by a multiple of two without affecting the greatest common divisor. Since the <math>n^2</math> term is quite restrictive, let's multipy by <math>4</math> so that we can get a <math>(2n+1)^2</math> in there.
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Thus the [[equation]] turns into <math>d_n=(100+n^2,2n+1)</math>. Now note that since <math>2n+1</math> is [[odd integer | odd]] for [[integer | integral]] <math>n</math>, we can multiply the left integer, <math>100+n^2</math>, by a multiple of two without affecting the greatest common divisor. Since the <math>n^2</math> term is quite restrictive, let's multipy by <math>4</math> so that we can get a <math>(2n+1)^2</math> in there.
  
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So <math>d_n=(4n^2+400,2n+1)=((2n+1)^2-4n+399,2n+1)=(-4n+399,2n+1)</math>. It simplified the way we wanted it to!
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So <math>\displaystyle d_n=(4n^2+400,2n+1)=((2n+1)^2-4n+399,2n+1)=(-4n+399,2n+1)</math>. It simplified the way we wanted it to!
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Now using similar techniques we can write <math>d_n=(-2(2n+1)+401,2n+1)=(401,2n+1)</math>. Thus the maximum value of <math>d_n</math> is <math>401</math>.
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Now using similar techniques we can write <math>d_n=(-2(2n+1)+401,2n+1)=(401,2n+1)</math>. Thus <math>d_n</math> must divide <math>401</math> for every single <math>n</math>.  This means the largest possible value for <math>d_n</math> is <math>401</math>, and we see that it can be achieved when <math>n = 200</math>.
  
  
  
 
== See also ==
 
== See also ==
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* [[1985 AIME Problems/Problem 14 | Next problem]]
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* [[1985 AIME Problems/Problem 12 | Previous problem]]
 
* [[1985 AIME Problems]]
 
* [[1985 AIME Problems]]
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 +
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[[Category:Intermediate Number Theory Problems]]

Revision as of 16:22, 12 October 2006

Problem

The numbers in the sequence $101$, $104$, $109$, $116$,$\ldots$ are of the form $a_n=100+n^2$, where $n=1,2,3,\ldots$ For each $n$, let $d_n$ be the greatest common divisor of $a_n$ and $a_{n+1}$. Find the maximum value of $d_n$ as $n$ ranges through the positive integers.

Solution

If $(x,y)$ denotes the greatest common divisor of $x$ and $y$, then we have $d_n=(a_n,a_{n+1})=(100+n^2,100+n^2+2n+1)$. Now assuming that $d_n$ divides $100+n^2$, it must divide $2n+1$ if it is going to divide the entire expression $100+n^2+2n+1$.

Thus the equation turns into $d_n=(100+n^2,2n+1)$. Now note that since $2n+1$ is odd for integral $n$, we can multiply the left integer, $100+n^2$, by a multiple of two without affecting the greatest common divisor. Since the $n^2$ term is quite restrictive, let's multipy by $4$ so that we can get a $(2n+1)^2$ in there.

So $\displaystyle d_n=(4n^2+400,2n+1)=((2n+1)^2-4n+399,2n+1)=(-4n+399,2n+1)$. It simplified the way we wanted it to! Now using similar techniques we can write $d_n=(-2(2n+1)+401,2n+1)=(401,2n+1)$. Thus $d_n$ must divide $401$ for every single $n$. This means the largest possible value for $d_n$ is $401$, and we see that it can be achieved when $n = 200$.


See also

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