1985 AIME Problems/Problem 13

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 $n = 200$ .

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
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1985 AIME Problems/Problem 13

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