Difference between revisions of "2009 AMC 12B Problems/Problem 19"
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Instead of doing the hard work, we can try to guess the factorization. One good approach: | 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 | + | 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. | 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. | ||
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== 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}} |
Revision as of 01:00, 11 January 2010
Problem
For each positive integer , let . What is the sum of all values of that are prime numbers?
Solution
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:
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 :
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.
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 |