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

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[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]

Revision as of 14:36, 6 May 2007

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

Thus the equation turns into $\displaystyle d_n=(100+n^2,2n+1)$. Now note that since $\displaystyle 2n+1$ is odd for integral $\displaystyle n$, we can multiply the left integer, $\displaystyle 100+n^2$, by a multiple of two without affecting the greatest common divisor. Since the $\displaystyle n^2$ term is quite restrictive, let's multiply by $\displaystyle 4$ so that we can get a $(\displaystyle 2n+1)^2$ 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 $\displaystyle d_n=(-2(2n+1)+401,2n+1)=(401,2n+1)$. Thus $\displaystyle d_n$ must divide $\displaystyle 401$ for every single $\displaystyle n$. This means the largest possible value for $\displaystyle d_n$ is $\displaystyle 401$, and we see that it can be achieved when $\displaystyle n = 200$.

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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