Difference between revisions of "2009 AMC 12B Problems/Problem 19"

Revision as of 13:30, 15 July 2015

Problem

For each positive integer $n$ , let $f(n) = n^4 - 360n^2 + 400$ . What is the sum of all values of $f(n)$ that are prime numbers?

$\textbf{(A)}\ 794\qquad \textbf{(B)}\ 796\qquad \textbf{(C)}\ 798\qquad \textbf{(D)}\ 800\qquad \textbf{(E)}\ 802$

Solution

Solution 1

To find the answer it was enough to play around with $f$ . One can easily find that $f(1)=41$ is a prime, then $f$ becomes negative for $n$ between $2$ and $18$ , and then $f(19)=761$ is again a prime number. And as $41 + 761 = 802$ is already the largest option, the answer must be $\boxed{802}$ .

Solution 2

We will now show a complete solution, with a proof that no other values are prime.

Consider the function $g(x) = x^2 - 360x + 400$ , then obviously $f(x) = g(x^2)$ .

The roots of $g$ are: $x_{1,2} = \frac{ 360 \pm \sqrt{ 360^2 - 4\cdot 400 } }2 = 180 \pm 80 \sqrt 5$

We can then write $g(x) = (x - 180 - 80\sqrt 5)(x - 180 + 80\sqrt 5)$ , and thus $f(x) = (x^2 - 180 - 80\sqrt 5)(x^2 - 180 + 80\sqrt 5)$ .

We would now like to factor the right hand side further, using the formula $(x^2 - y^2) = (x-y)(x+y)$ . To do this, we need to express both constants as squares of some other constants. Luckily, we have a pretty good idea how they look like.

We are looking for rational $a$ and $b$ such that $(a+b\sqrt 5)^2 = 180 + 80\sqrt 5$ . Expanding the left hand side and comparing coefficients, we get $ab=40$ and $a^2 + 5b^2 = 180$ . We can easily guess (or compute) the solution $a=10$ , $b=4$ .

Hence $180 + 80\sqrt 5 = (10 + 4\sqrt 5)^2$ , and we can easily verify that also $180 - 80\sqrt 5 = (10 - 4\sqrt 5)^2$ .

We now know the complete factorization of $f(x)$ :

$f(x) = (x - 10 - 4\sqrt 5)(x + 10 + 4\sqrt 5)(x - 10 + 4\sqrt 5)(x + 10 - 4\sqrt 5)$

As the final step, we can now combine the factors in a different way, in order to get rid of the square roots.

We have $(x - 10 - 4\sqrt 5)(x - 10 + 4\sqrt 5) = (x-10)^2 - (4\sqrt 5)^2 = x^2 - 20x + 20$ , and $(x + 10 - 4\sqrt 5)(x + 10 + 4\sqrt 5) = x^2 + 20x + 20$ .

Hence we obtain the factorization $f(x) = (x^2 - 20x + 20)(x^2 + 20x + 20)$ .

For $x\geq 20$ both terms are positive and larger than one, hence $f(x)$ is not prime. For $1<x<19$ the second factor is positive and the first one is negative, hence $f(x)$ is not a prime. The remaining cases are $x=1$ and $x=19$ . In both cases, $f(x)$ is indeed a prime, and their sum is $f(1) + f(19) = 41 + 761 = \boxed{802}$ .

Solution 3

Instead of doing the hard work, we can try to guess the factorization. One good approach:

We can make the observation that $f(x)$ looks similar to $(x^2 + 20)^2$ with the exception of the $x^2$ term. In fact, we have $(x^2 + 20)^2 = x^4 + 40x^2 + 400$ . But then we notice that it differs from the desired expression by a square: $f(x) = (x^2 + 20)^2 - 400x^2 = (x^2 + 20)^2 - (20x)^2$ .

Now we can use the formula $(x^2 - y^2) = (x-y)(x+y)$ to obtain the same factorization as in the previous solution, without all the work.

Solution 4

After arriving at the factorization $f(x) = (x^2 - 20x + 20)(x^2 + 20x + 20)$ , a more mathematical approach would be to realize that the second factor is always positive when $x$ is a positive integer. Therefore, in order for $f(x)$ to be prime, the first factor has to be $1$ .

We can set it equal to 1 and solve for $x$ :

$x^2-20x+20=1$

$x^2-20x+19=0$

$(x-1)(x-19)=0$

$x=1, x=19$

Substituting these values into the second factor and adding would give the answer.

Solution 5 (Using the answer choices)

Plugging in a few values, it is obvious that $f(1) = 41$ is prime. Starting from $n = 2$ , we know that $f(n)$ becomes negative, then positive again. We want the find the first integral $n$ in which $f(n)$ is positive. This is solved by $n^4 -360n^2+400>0$ .

Instead of using the quadratic formula, we can ignore $400$ to get an approximation. So, we want to solve $n^4-360n^2>0$ . Simplifying, $n^2>360$ . Looking at this, we can automatically see that $n = 19$ is the first value that this happens.

We are only looking for odd values of $n$ . Although we can check if $n = 18$ is positive to account for the $400$ we took off, this will be unnecessary because it wouldn’t be prime anyways.

Plugging in $n = 19$ , we get $f(19) = 19^2(19^2-360)+400 = 19^2*1+400 = 761$ . Since we know that $41+761 = 802$ and $802$ is the largest answer choice, it follows that the answer is $\text{E }\boxed{802}$ .

@@ Line 81: / Line 81: @@
 Plugging in a few values, it is obvious that <math>f(1) = 41</math> is prime. Starting from <math>n = 2</math>, we know that <math>f(n)</math> becomes negative, then positive again. We want the find the first integral <math>n</math> in which <math>f(n)</math> is positive. This is solved by <math>n^4 -360n^2+400>0</math>.
-Instead of using the quadratic formula, we can ignore <math>400</math> to get an approximation. So, we want to solve<math>n^4-360n^2>0</math>.
+Instead of using the quadratic formula, we can ignore <math>400</math> to get an approximation. So, we want to solve <math>n^4-360n^2>0</math>.
 Simplifying, <math>n^2>360</math>. Looking at this, we can automatically see that <math>n = 19</math> is the first value that this happens.
 We are only looking for odd values of <math>n</math>. Although we can check if <math>n = 18</math> is positive to account for the <math>400</math> we took off, this will be unnecessary because it wouldn’t be prime anyways.
 Plugging in <math>n = 19</math>, we get <math>f(19) = 19^2(19^2-360)+400 = 19^2*1+400 = 761</math>. Since we know that <math>41+761 = 802</math> and <math>802</math> is the largest answer choice, it follows that the answer is <math>\text{E }\boxed{802}</math>.

2009 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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