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

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== Solution ==
 
== Solution ==
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If <math>\displaystyle (x,y)</math> denotes the [[greatest common divisor]] of <math>\displaystyle x</math> and <math>\displaystyle y</math>, then we have <math>\displaystyle d_n=(a_n,a_{n+1})=(100+n^2,100+n^2+2n+1)</math>. Now assuming that <math>\displaystyle d_n</math> [[divisor | divides]] <math>\displaystyle 100+n^2</math>, it must divide <math>\displaystyle 2n+1</math> if it is going to divide the entire [[expression]] <math>\displaystyle 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>\displaystyle d_n=(100+n^2,2n+1)</math>. Now note that since <math>\displaystyle 2n+1</math> is [[odd integer | odd]] for [[integer | integral]] <math>\displaystyle n</math>, we can multiply the left integer, <math>\displaystyle 100+n^2</math>, by a multiple of two without affecting the greatest common divisor. Since the <math>\displaystyle n^2</math> term is quite restrictive, let's multiply by <math>\displaystyle 4</math> so that we can get a <math>(\displaystyle 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 multiply by <math>4</math> so that we can get a <math>(2n+1)^2</math> in there.
  
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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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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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Now using similar techniques we can write <math>\displaystyle d_n=(-2(2n+1)+401,2n+1)=(401,2n+1)</math>.  Thus <math>\displaystyle d_n</math> must divide <math>\displaystyle 401</math> for every single <math>\displaystyle n</math>.  This means the largest possible value for <math>\displaystyle d_n</math> is <math>\displaystyle 401</math>, and we see that it can be achieved when <math>\displaystyle n = 200</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>\boxed{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 ==

Revision as of 21:34, 20 September 2011

Problem

The numbers in the sequence $\displaystyle 101$, $\displaystyle 104$, $\displaystyle 109$, $\displaystyle 116$,$\displaystyle \ldots$ are of the form $\displaystyle a_n=100+n^2$, where $\displaystyle n=1,2,3,\ldots$ For each $\displaystyle n$, let $\displaystyle d_n$ be the greatest common divisor of $\displaystyle a_n$ and $\displaystyle a_{n+1}$. Find the maximum value of $\displaystyle d_n$ as $\displaystyle n$ 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 multiply by $4$ so that we can get a $(2n+1)^2$ in there.

So $d_n=(4n^2+400,2n+1)=((2n+1)^2-4n+399,2n+1)=(-4n+399,2n+1)$. 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 $\boxed{401}$ 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

1985 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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