Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(New page: == Problem == For each positive integer <math>n</math>, let <math>f(n) = n^4 - 360n^2 + 400</math>. What is the sum of all values of <math>f(n)</math> that are prime numbers? <math>\textb...)
 
(→‎Solution)
 
(20 intermediate revisions by 6 users not shown)
Line 4: Line 4:
 
<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>
  
−
== Solution ==
+
== Solutions ==
  
 
=== Solution 1 ===
 
=== Solution 1 ===
Line 23: Line 23:
 
</cmath>
 
</cmath>
  
−
We can then write <math>g(x) = (x - 180 - 80\sqrt 5)(x - 180 - 80\sqrt 5)</math>, and thus <math>f(x) = (x^2 - 180 - 80\sqrt 5)(x^2 - 180 - 80\sqrt 5)</math>.
+
We can then write <math>g(x) = (x - 180 - 80\sqrt 5)(x - 180 + 80\sqrt 5)</math>, and thus <math>f(x) = (x^2 - 180 - 80\sqrt 5)(x^2 - 180 + 80\sqrt 5)</math>.
  
 
We would now like to factor the right hand side further, using the formula <math>(x^2 - y^2) = (x-y)(x+y)</math>. 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 would now like to factor the right hand side further, using the formula <math>(x^2 - y^2) = (x-y)(x+y)</math>. 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.
Line 39: Line 39:
 
As the final step, we can now combine the factors in a different way, in order to get rid of the square roots.  
 
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 <math>(x - 10 - 4\sqrt 5)(x - 10 + 4\sqrt 5) = (x-10)^2 - (4\sqrt 5)^2 = (x^2 - 20x + 100) - 80 = x^2 - 20x + 20</math>,
+
We have <math>(x - 10 - 4\sqrt 5)(x - 10 + 4\sqrt 5) = (x-10)^2 - (4\sqrt 5)^2 = x^2 - 20x + 20</math>,
 
and <math>(x + 10 - 4\sqrt 5)(x + 10 + 4\sqrt 5) = x^2 + 20x + 20</math>.
 
and <math>(x + 10 - 4\sqrt 5)(x + 10 + 4\sqrt 5) = x^2 + 20x + 20</math>.
  
Line 45: Line 45:
  
 
For <math>x\geq 20</math> both terms are positive and larger than one, hence <math>f(x)</math> is not prime. For <math>1<x<19</math> the second factor is positive and the first one is negative, hence <math>f(x)</math> is not a prime. The remaining cases are <math>x=1</math> and <math>x=19</math>. In both cases, <math>f(x)</math> is indeed a prime, and their sum is <math>f(1) + f(19) = 41 + 761 = \boxed{802}</math>.
 
For <math>x\geq 20</math> both terms are positive and larger than one, hence <math>f(x)</math> is not prime. For <math>1<x<19</math> the second factor is positive and the first one is negative, hence <math>f(x)</math> is not a prime. The remaining cases are <math>x=1</math> and <math>x=19</math>. In both cases, <math>f(x)</math> is indeed a prime, and their sum is <math>f(1) + f(19) = 41 + 761 = \boxed{802}</math>.
 +
 +
=== Solution 3 ===
 +
 +
Instead of doing the hard work, we can try to guess the factorization. One good approach:
 +
 +
We can make the observation that <math>f(x)</math> looks similar to <math>(x^2 + 20)^2</math> with the exception of the <math>x^2</math> term. In fact, we have <math>(x^2 + 20)^2 = x^4 + 40x^2 + 400</math>. But then we notice that it differs from the desired expression by a square: <math>f(x) = (x^2 + 20)^2 - 400x^2 = (x^2 + 20)^2 - (20x)^2</math>.
 +
 +
Now we can use the formula <math>(x^2 - y^2) = (x-y)(x+y)</math> to obtain the same factorization as in the previous solution, without all the work.
 +
 +
=== Solution 4 ===
 +
 +
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>:
 +
 +
<math>
 +
x^2-20x+20=1
 +
</math>
 +
 +
<math>
 +
x^2-20x+19=0
 +
</math>
 +
 +
<math>
 +
(x-1)(x-19)=0
 +
</math>
 +
 +
<math>
 +
x=1, x=19
 +
</math>
 +
 +
Substituting these values into the second factor and adding would give the answer.
  
 
== 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}}
 +
{{MAA Notice}}

Latest revision as of 13:27, 23 December 2020

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.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_19&oldid=140372"