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> | ||
| − | == | + | == Solutions == |
=== Solution 1 === | === Solution 1 === | ||
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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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== See Also == | == See Also == | ||
Revision as of 12:27, 23 December 2020
Contents
Problem
For each positive integer 
, let 
. What is the sum of all values of 
that are prime numbers?
Solutions
Solution 1
To find the answer it was enough to play around with 
. One can easily find that 
is a prime, then becomes negative for between 
and 
, and then 
is again a prime number. And as 
is already the largest option, the answer must be 
.
Solution 2
We will now show a complete solution, with a proof that no other values are prime.
Consider the function 
, then obviously 
.
The roots of 
are:
![\[x_{1,2} = \frac{ 360 \pm \sqrt{ 360^2 - 4\cdot 400 } }2 = 180 \pm 80 \sqrt 5\]](http://latex.artofproblemsolving.com/b/d/6/bd611e70153e15741f627754fda139a655bfc44a.png)
We can then write 
, and thus 
.
We would now like to factor the right hand side further, using the formula 
. 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 
and 
such that 
. Expanding the left hand side and comparing coefficients, we get 
and 
. We can easily guess (or compute) the solution 
, 
.
Hence 
, and we can easily verify that also 
.
We now know the complete factorization of 
:
![\[f(x) = (x - 10 - 4\sqrt 5)(x + 10 + 4\sqrt 5)(x - 10 + 4\sqrt 5)(x + 10 - 4\sqrt 5)\]](http://latex.artofproblemsolving.com/e/6/5/e65b2163ba192ddf7afcfe37167b941d980c53a3.png)
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 
,
and 
.
Hence we obtain the factorization 
.
For 
both terms are positive and larger than one, hence is not prime. For 
the second factor is positive and the first one is negative, hence is not a prime. The remaining cases are 
and 
. In both cases, is indeed a prime, and their sum is 
.
Solution 3
Instead of doing the hard work, we can try to guess the factorization. One good approach:
We can make the observation that looks similar to 
with the exception of the 
term. In fact, we have 
. But then we notice that it differs from the desired expression by a square: 
.
Now we can use the formula to obtain the same factorization as in the previous solution, without all the work.
Solution 4
After arriving at the factorization , a more mathematical approach would be to realize that the second factor is always positive when 
is a positive integer. Therefore, in order for to be prime, the first factor has to be 
.
We can set it equal to 1 and solve for :
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. 
