1985 AIME Problems/Problem 13

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 $\displaystyle (x,y)$ 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 $\displaystyle n = 200$ .

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1985 AIME Problems/Problem 13

Problem

Solution

See also