2009 AMC 12B Problems/Problem 19
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 |
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All AMC 12 Problems and Solutions |