AIME 1985 · 第 13 题

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 power 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 $100+n^2$.

$$(100+n^2)-100(2n+1)$$ $$=n^2+100-200n-100$$ This leaves us with

$$n^2-200n.$$ Factoring, we get

$$n(n-200)$$ Because $n$ and $2n+1$ will be coprime, the only thing stopping the GCD from being $1$ is $n-200.$ We want this to equal 0, because that will maximize our GCD. 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, $\gcd(x-200,2x+1)=\gcd(x-200,2x+1-2(x-200))=\gcd(x-200,401)$. 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 $\boxed{401}$, and happens when $n \equiv 200 \pmod{401}$.

Solution 5

As clearly shown in the above solutions, we want to maximize $(100+n^2, 2n+1).$ Then the maximum of the GCD will be achieved with integers $a,b,c$ such that $a(100+n^2) + b(2n+1)=c$ by Bezout's identity. Note for the LHS to be constant, $b$ must be a linear function of $n,$ so $b=px+q.$ However, there cannot be a linear $n$ term in $b(2n+1),$ hence $b=2n-1$ by difference of squares. Changing $b$ to $-2n+1,$ we get $100a+an^2-4n^2+1=c,$ so $a=4,$ and the answer is $c=\boxed{401}.$

~anduran

Video Solution by OmegaLearn

https://youtu.be/yh70NBCxQzg?t=752

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