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

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<math>\textbf{(A)}\ 794\qquad \textbf{(B)}\ 796\qquad \textbf{(C)}\ 798\qquad \textbf{(D)}\ 800\qquad \textbf{(E)}\ 802</math>
 
<math>\textbf{(A)}\ 794\qquad \textbf{(B)}\ 796\qquad \textbf{(C)}\ 798\qquad \textbf{(D)}\ 800\qquad \textbf{(E)}\ 802</math>
  
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== Solution ==
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== Solutions ==
  
 
=== Solution 1 ===
 
=== Solution 1 ===
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=== Solution 4 ===
 
=== Solution 4 ===
  
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After arriving at the factorization <math>f(x) = (x^2 - 20x + 20)(x^2 + 20x + 20)</math>, 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 <math>f(x)</math> to be prime, the first factor has to be <math>1</math>.
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After arriving at the factorization <math>f(x) = (x^2 - 20x + 20)(x^2 + 20x + 20)</math>, a more mathematical approach would be to realize that the second factor is always positive when <math>x</math> is a positive integer. Therefore, in order for <math>f(x)</math> to be prime, the first factor has to be <math>1</math>.
  
 
We can set it equal to 1 and solve for <math>x</math>:  
 
We can set it equal to 1 and solve for <math>x</math>:  
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Substituting these values into the second factor and adding would give the answer.
 
Substituting these values into the second factor and adding would give the answer.
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=== A way to find the desired factorization ===
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Essentially since we want to find when the function is prime, we shall attempt to factor the polynomial. Notice we have <math>(n^2+20)^2-400n^2=n^4-360n^2+400</math>, and difference of squares gives the desired factorization, and we can continue as in any other solution. Furthermore, the motivation from doing <math>(n^2+20)^2</math> is since it matches the leading term in f(x) and the constant term.
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2009|ab=B|num-b=18|num-a=20}}
 
{{AMC12 box|year=2009|ab=B|num-b=18|num-a=20}}
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{{MAA Notice}}

Latest revision as of 12:28, 25 September 2024

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?

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

Solutions

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.

A way to find the desired factorization

Essentially since we want to find when the function is prime, we shall attempt to factor the polynomial. Notice we have $(n^2+20)^2-400n^2=n^4-360n^2+400$, and difference of squares gives the desired factorization, and we can continue as in any other solution. Furthermore, the motivation from doing $(n^2+20)^2$ is since it matches the leading term in f(x) and the constant term.

See Also

2009 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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