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

Revision as of 16:35, 1 January 2018

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 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.

Solution 4

We can just plug in Euclidean algorithm, to go from $gcd(n^2 + 100, n^2 + 2n + 101)$ to $gcd(n^2 + 100, 2n + 1)$ to $gcd(n^2 + 100 - 100(2n + 1), 2n + 1)$ to get $gcd(n^2 - 200n, 2n + 1)$ . Now we know that no matter what, $n$ is relatively prime to $2n + 1$ . Therefore the equation can be simplified to: $gcd(n - 200, 2n + 1)$ . Subtracting $2n - 400$ from $2n + 1$ results in $gcd(n - 200,401)$ . The greatest possible value of this is $\boxed{401}$ , an happens when $n \equiv 200 \mod{401}$ .

@@ Line 1: / Line 1: @@
 == Problem ==
-The numbers in the [[sequence]] <math>\displaystyle 101</math>, <math>\displaystyle 104</math>, <math>\displaystyle 109</math>, <math>\displaystyle 116</math>,<math>\displaystyle \ldots</math> are of the form <math>\displaystyle a_n=100+n^2</math>, where <math>\displaystyle n=1,2,3,\ldots</math> For each <math>\displaystyle n</math>, let <math>\displaystyle d_n</math> be the greatest common divisor of <math>\displaystyle a_n</math> and <math>\displaystyle a_{n+1}</math>. Find the maximum value of <math>\displaystyle d_n</math> as <math>\displaystyle n</math> ranges through the [[positive integer]]s.
+The numbers in the [[sequence]] <math>101</math>, <math>104</math>, <math>109</math>, <math>116</math>,<math>\ldots</math> are of the form <math>a_n=100+n^2</math>, where <math>n=1,2,3,\ldots</math> For each <math>n</math>, let <math>d_n</math> be the greatest common divisor of <math>a_n</math> and <math>a_{n+1}</math>. Find the maximum value of <math>d_n</math> as <math>n</math> ranges through the [[positive integer]]s.
 == Solution 1==
@@ Line 19: / Line 19: @@
 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.
+==Solution 4==
+We can just plug in Euclidean algorithm, to go from <math>gcd(n^2 + 100, n^2 + 2n + 101)</math> to <math>gcd(n^2 + 100, 2n + 1)</math> to <math>gcd(n^2 + 100 - 100(2n + 1), 2n + 1)</math> to get <math>gcd(n^2 - 200n, 2n + 1)</math>. Now we know that no matter what, <math>n</math> is relatively prime to <math>2n + 1</math>. Therefore the equation can be simplified to: <math>gcd(n - 200, 2n + 1)</math>. Subtracting <math>2n - 400</math> from <math>2n + 1</math> results in <math>gcd(n - 200,401)</math>. The greatest possible value of this is <math>\boxed{401}</math>, an happens when <math>n \equiv 200 \mod{401}</math>.
 == See also ==
 {{AIME box|year=1985|num-b=12|num-a=14}}

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

Page

Toolbox

Search