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

Revision as of 14:51, 19 September 2015

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 1

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 $n = 200$ .

Solution 2

We know that $a_n = 100+n^2$ and $a_{n+1} = 100+(n+1)^2 = 100+ n^2+2n+1$ . Since we want to find the GCD of $a_n$ and $a_{n+1}$ , we can use the Euclidean algorithm:

$a_{n+1}-a_n = 2n+1$

Now, the question is to find the GCD of $2n+1$ and $100+n^2$ . We subtract $2n+1$ 100 times from $100+n^2$ . This leaves us with $n^2-200n$ . We want this to equal 0, so solving for $n$ gives us $n=200$ . The last remainder is 0, thus $200*2+1 = \boxed{401}$ is our GCD.

Solution 3

If solution 2 is not entirely obvious, our answer is the max possible range of $\frac{x(x-200)}{2x+1}$ . Using the euclidean algorithm on $x$ and $2x+1$ yields that they are relatively prime. Thus, the only way the gcd will not be 1 is if the $x-200$ term share factors with the $2x+1$ . Using the euclidean algorithm, $gcd(x-200,2x+1)=gcd(x-22,2x+1-2(x-200))=gcd(x-200,401)$ . Thus, the max gcd is 401.

@@ Line 17: / Line 17: @@
 Now, the question is to find the GCD of <math>2n+1</math> and <math>100+n^2</math>. We subtract <math>2n+1</math> 100 times from <math>100+n^2</math>. This leaves us with <math>n^2-200n</math>. We want this to equal 0, so solving for <math>n</math> gives us <math>n=200</math>. The last remainder is 0, thus <math>200*2+1 = \boxed{401}</math> is our GCD.
 == Solution 3==
-If solution 2 is not entirely obvious, our answer is the max possible range of <math>/frac{x(x-200)}{2x+1}</math>. Using the euclidean algorithm on <math>x</math> and <math>2x+1</math> yields that they are relatively prime. Thus, the only way the gcd will not be 1 is if the<math> x-200</math> term share factors with the <math>2x+1</math>. Using the euclidean algorithm, <math>gcd(x-200,2x+1)=gcd(x-22,2x+1-2(x-200))=gcd(x-200,401)</math>. Thus, the max gcd is 401.
+If solution 2 is not entirely obvious, our answer is the max possible range of <math>\frac{x(x-200)}{2x+1}</math>. Using the euclidean algorithm on <math>x</math> and <math>2x+1</math> yields that they are relatively prime. Thus, the only way the gcd will not be 1 is if the<math> x-200</math> term share factors with the <math>2x+1</math>. Using the euclidean algorithm, <math>gcd(x-200,2x+1)=gcd(x-22,2x+1-2(x-200))=gcd(x-200,401)</math>. Thus, the max gcd is 401.
 == 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
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